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If 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

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βˆ™ 15y ago
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βˆ™ 1mo ago

You 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.

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βˆ™ 14y ago

If 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.

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Q: How do you find time of flight from initial velocity and an angle at which an object is thrown?
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