Wiki User
β 14y agoIf you're willing to ignore the effect of air resistance, then the answer is as follows: The object's horizontal velocity remains constant (at least until it eventually hits the ground). The vertical component of the object's initial velocity ... call it V(i) ... is the (total initial velocity) multipled by the (sine of the initial angle above the horizontal). Beginning at the time of the toss, the magnitude of the vertical component of velocity is V = V(i) - 1/2gT2. T = number of seconds after the toss g = acceleration of gravity = approx 32 ft/sec2 or 9.8 m/sec2
Wiki User
β 15y agoYou can find the time of flight by using the formula: time of flight = (2 * initial velocity * sin(angle)) / gravitational acceleration. Input the initial velocity and angle at which the object is thrown into the formula to calculate the time it takes for the object to reach the same height as it was initially launched.
Wiki User
β 14y agoIf you have the initial velocity of an object and the angle at which it was thrown, you can calculate the time of flight.
First, you must break the velocity into its x and y components. The x component can be found by using the equation:
Vcos(theta) = x component of the velocity
(V is the velocity and theta represents the angle at which the object was thrown.)
The y component can be found with:
Vsin(theta) = y component of the velocity
In this problem, we only need to worry about the y component. Once you calculate it, you can plug it into this equation:
t = g/v
(since this is in the y direction, the acceleration will be g, gravity or 9.8 m/s^2)
This will give you the time it takes the object to reach it's peak height. It will take the same amount of time to then fall to the ground, so the final answer will be 2t.
The path that a thrown object follows is called a projectile path, which is determined by the initial velocity and angle of the throw. The object moves in a curved trajectory, influenced by gravity pulling it downward. The shape of this path is typically a parabolic curve.
In order for a body to escape the gravitational pull of the Earth, it needs to be thrown up with an initial velocity equal to or greater than the escape velocity of around 11.2 km/s. This velocity allows the object to overcome the gravitational pull of the Earth and continue traveling away from it indefinitely.
The velocity with which the object is thrown upwards can be found using the equation v = u + at, where v is the final velocity (0 m/s at the top), u is the initial velocity, a is the acceleration due to gravity (-9.81 m/s^2), and t is the time taken to reach the ground (4 seconds). Rearranging the equation to solve for u, we have u = v - at. Plugging in the values, u = 0 - (-9.81 * 4) = 39.24 m/s. Therefore, the object is thrown upwards with a velocity of 39.24 m/s.
The acceleration of an object thrown vertically upwards can be calculated using the kinematic equation (v_f^2 = v_i^2 + 2a \cdot d), where (v_f) is the final velocity, (v_i) is the initial velocity, (a) is the acceleration, and (d) is the distance. Given that the object is thrown vertically upwards, the equation becomes (0 = (44 , \text{m/s})^2 + 2 \cdot a \cdot (-3.5 , \text{m})). Solving for (a), we find that the acceleration is approximately -104 m/sΒ², which indicates that the object is accelerating downwards.
The impact velocity of a rock thrown horizontally from a cliff will depend on the height of the cliff and the initial velocity at which it was thrown. If air resistance is neglected, the velocity of impact will be equal to the horizontal component of the initial velocity.
The best way to analyze an object thrown into the air is to use projectile motion equations to calculate its initial velocity, angle of projection, and other relevant parameters. By breaking down the motion into horizontal and vertical components, you can determine its trajectory, maximum height, range, and time of flight accurately. Additionally, considering air resistance if necessary can provide a more realistic analysis of the object's motion.
Ignoring air resistance, I get this formula:Maximum height of a vertically-launched object = 1.5 square of initial speed/GI could be wrong. In that case, the unused portion of my fee will be cheerfully refunded.
No, the acceleration is not the same for an object that is dropped and an object that is thrown. When an object is dropped, it experiences a constant acceleration due to gravity. When an object is thrown, its acceleration can vary depending on factors such as the initial velocity and direction.
Yes, it is possible for the initial velocity to be different from zero when the final velocity is zero. For example, an object could be thrown upwards and come to a stop at its highest point, where the final velocity would be zero.
Objects that are thrown follow a curved path due to the influence of gravity and the initial forward velocity given to the object. Gravity pulls the object downward, causing it to accelerate, while the initial forward velocity allows the object to travel horizontally. The combination of these two forces results in the object following a curved trajectory known as a parabola.
The vertical component of the initial velocity of the ball thrown horizontally from a window is zero. The ball's initial velocity in the vertical direction is influenced only by the force of gravity, not the horizontal throw.
Acceleration is dependent on the initial velocity of how fast the object is leaving the projectile. The vertical acceleration is greater when the object is falling than when the object reaches the peak in height. However, if the object is thrown horizontally and there is no parabola in its shape then there is not as great of an acceleration.
30 mph!
Gravity acts the same way on objects falling freely down and those thrown upwards. The difference lies in the initial velocity and direction of the objects. Objects thrown upwards have an initial velocity that opposes gravity, causing them to slow down and eventually fall back down due to gravity. Objects falling freely down have an initial velocity of zero and accelerate towards the ground due to gravity.
The motion of an object thrown at an angle is a combination of both horizontal and vertical motion. The horizontal motion is constant and is controlled by the initial velocity in the x-direction. The vertical motion is influenced by gravity, causing the object to follow a curved path.
The motion of an object thrown at an angle is projectile motion. This type of motion involves the object following a curved path due to a combination of its initial velocity and the force of gravity acting upon it. The object moves both horizontally and vertically as it travels through the air.
If the height from which the ball is thrown is increased, the time of flight of the ball would increase as well. This is because the initial velocity of the ball would be higher, leading to a longer time for the ball to reach the ground.