Question

A tennis ball is thrown towards a tall wall that is 20 m away. The initial velocity of the ball is 10 m/s at 60o above the horizontal. The ball will hit the wall what time after release?

Answer 1

Answer:

Time=3.1 seconds

Explanation:

Given data

Distance S=20 m

Velocity v=10×Sin60°

To find

Time after release

Solution

From equation of simple motion we know that

$S=vt+(1/2)gt^{2}$

Using the convention that down is positive then

$20m=-10sin60^{o}t+(1/2)(9.8m/s^{2} )t^{2}\\ 20=-8.66t+4.9t^{2}\\ -4.9t^{2}+8.66t+20=0$

Apply Quadratic formula  to solve t

$t=\frac{-(8.66)+\sqrt{(8.66)^{2}+4(-4.9)(20) } }{2(-4.9)}$

Ignoring the -ve root then we have

t=3.1 seconds