Question

A third baseman makes a throw to first base 40 m away. The ball leaves his hand with a speed of 37 m/s at a height of 1.5 m from the ground and making an angle of 19 degrees with the horizontal. How high will the ball be when it gets to first base?

Answer 1

Hello,

Third baseman throws the ball at a speed of 37m/s at an angle of 19deg from 1.5m high. Hence the path ball takes is of parabolic nature.

The only force acting on the ball in air is gravity acting downwards.

The speed can be divided into two components one horizontal and one vertical.

Horizontal speed component = 37*cos(19)=34.98m/s

Vertical speed component=37*sin(19)=12.04m/s

There is no force acting horizontally on the body. Hence the ball's motion is that of uniform velocity along the horizontal axis without any acceleration.(No force implies no acceleration as F=ma and no acceleration implies no change in velocity).

The time taken for the ball to advance a horizontal distance of 40m is given by t=40/34.98=1.14sec.

Now our aim is to look at how much high the ball wouldve gone in that time of 1.14sec.

The vertical component of speed experiences deceleration due to gravity acting downwards. Therefore its magnitude keeps decreasing as the ball progresses in height. The vertical distance travelled by ball in 1.14sec is therefore given by

s=u*t+1/2*a*t²

==> 12.04*1.14+1/2*(-9.8)*1.14²

==> 7.35m

Since the ball was already 1.5m high when thrown into air, total height when first base receives it is 7.35+1.5=8.85m