Question

Jose is batting for the home team when he hits a foul ball that rises straight up over home plate. A fan in the stands notices the ball reaches the same height as the top row of the stands where he is sitting. If the top row of the stands is 40 meters above the field level, what was the upward speed of the ball in meters per second after it got struck by the bat?

Answer 1

28.01m/s.

Step-by-step explanation:

Given maximum height reached by the ball as H = 40 metres

Since the ball rises straight up when hit by a ball, then the angle of launch will be perpendicular to the ground and that is 90°.

To determine the upward speed of the ball in meters per second after it got struck by the bat, we will use the formula for calculating the maximum height according to projectile motion;

Maximum Height H =
`(u^2sin^2\theta)/(2g)` where;

u is the speed of the ball

`\theta` is the angle of launch

g is the acceleration due to gravity = 9.81m/s²

Substituting the given parameters into the formula;

`40 = (u^2sin^2(90))/(2(9.81))\\ \\40 = (u^2)/(2(9.81))\\ \\40 = (u^2)/(19.62) \\cross\ multiply\\\\u^2 = 40*19.62\\u^2 = 784.8\\u = √(784.8)\\ u = 28.01 m/s`

Hence the upward speed of the ball in meters per second after it got struck by the bat is 28.01m/s.