Question

An archer shoots an arrow at 83.0 m/s at a 62.0 degree angle. If the ground is flat, how much time is the arrow in the air?

Answer 1

The time taken is t = 14.96 s

Explanation:

In order to find the time of flight of the arrow, we just need to analyze its vertical motion. We can do it by using the following equation of motion:

Vy = uy + at ---- (1)

Where Vy is the vertical velocity at time "t".

uy is the initial vertical velocity.

a = g = -9.8 m/s^2

The arrow reaches its highest point in the trajectory when the vertical velocity, so when

vy=0

Also, the initial vertical velocity is given by

uy = u sin (θ) = (83.0)(sin 62.0°) = 73.3 m/s

Therefore, from (1) we find

Therefore, from (1) we find

t = - uy / g

t = -73.3 / -9.8

t = 7.48 s

This is the time the arrow needs to reach the highest point: the total time of flight of the arrow is just twice this time, so

T = 2t = 14.96 s

Hence the time taken is t =  14.96 s

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