How long is ball in air and tallest spot in the balls path

Question

A baseball is thrown at an angle of 23? relative to the ground at a speed of 24.2 m/s.The ball is caught 42.9 m from the thrower.
The acceleration of gravity is 9.81 m/s2.
a) How long is it in the air?
Answer in units of s.

 

b) How high is the tallest spot in the ball’s
path?
Answer in units of m

Answer 1

Let the vertical component of the speed to be V and horizontal component of the speed to be H.

 

Distance = H*t

42.9 = 24.2 * cos23 * t

t = 42.9 /(24.2*cos23) = 1.9268 sec   --- the time the ball is in the air

 

 

Height = V*t - 0.5*9.8*t^2

If we allow the ball to free fall to the ground without catching it, it will take:  

0 = 24.2 * Sin23 * t -0.5 * 9.8 * t^2          (where 23 is in degrees)

t = 1.929 sec ;

The ball was catch split second before it hits the ground, and the highest point it reaches is at t = 0.5 * 1.929 = 0.9645 sec

 

Height = 24.2 * Sin23 * 0.9645 -0.5 * 9.8 * (0.9645)^2

= 9.126 - 4.56 = 4.563 m