Question

A tennis ball is hit with a vertical speed of 35 m/s and a horizontal speed of 20 m/s. How long will the ball remain in the air?

Answer 1

Given

the speeds of the tennis ball hit with an angle of projection is theta is

Vy = 35 m/s, Vx = 20 m/s

===> the initial speed is v = sqrt(vx^2+vy^2)

v = sqrt(35^2+20^2) m/s

v = 40.311 m/s

we know that in projectile motion the horizontal velocity does not change but the vertical velcoty will

so

20 = 40.311 cos theta

==> theta = 60.25 degrees

How long will the ball remain in the air means the time of flight of the ball to reach the same point of hit

we know in projectile the time of flight is

T = 2*u sin theta/g

T = 2*40.311 sin 60.25 /9.8

T = 7.142 s