A batter hits a baseball so that it leaves the bat with an initial speed of 60m/s at an initial angle of 42 with the horizontal Find: The position of the ball and the magnitude and direction of its velocity at t = 2s Find the time when the ball reaches the highest point of its flight Find the height H and the acceleration at this point Find the horizontal range H.

24.06°

Explanation:

So,we can say after t if it reaches height h then,

h = (37 sin 53)t - 1/2 * 9.8t^2 (as,vertical component of velocity is 37 sin 53)

Given t = 2s

So, h = 39.5m

And horizontal displacement will be

r = 37 cos 53 *2 = 44.52m

So,after 2s the baseball will be lying 39.57m

above its point of projection and 44.52m ahead of its point of projection.

Now.let the vertical component of velocity will become Vy after time 2s

So, Uy = 37 sin 53- 9.8* 2

or, U = 9.95m/s

And.horizontal component of velocity remains

constant i.e Vx = 37 cos 53 = 22.27m/s

So.magnitude of velocity after 2s is

Square root of (Vx^2 + Vy^2)= 24.4m/s

Making an angle of tan 22.27/9.95 = 24.06°

3.2x10^7

the answer is a. hope this .