Question

A baseball player hits a home run, and the baseball just clear a
wall 27 m located 144 m from the home plate. The ball is hit at an
angle of 38 degrees to the horizontal, and air resistance is
negligible. Assume the ball is hit at a height of 1.0 m above the
ground. What is the initial speed of the ball? (use g = 10
m/s^{2}).

How much time does it take for the ball to reach the wall?

Find the components of the velocity and the speed of the ball as it reaches the wall.

What is the maximum height the ball reaches?

Answer #1

For horizontal motion,

144 = u cos38 x t

ut = 144/ cos38 = 182.74 (i)

For vertical motion,

h = ho + (u sin(theta))t - 0.5gt^2

1 = 27 + ut sin38 - 4.9t^2 = 27 + 182.74 x 0.62 - 5*t^2

t = 5.263 sec

Putting in (i)

u= 182.74/5.263= 34.72 m/s

======

b)

t= 5.263 seconds

======

c)

Vx= 34.72 cos 38= 27.36 m/s

Vy= 34.72 sin 38 - 10* 5.26= -31.22 m/s

Net velocity

V^2= Vx^2 + Vy^2

V= 41.515 m/s

=====

Max height

H= 1+ ( 31.22^2)/(2*10)= 49.734 m from ground level

======

Comment in case any doubt.. good luck

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