A baseball is thrown at an angle of 24◦ relative to the ground at a speed of 25.1 m/s. The ball
is caught 47.7257 m from the thrower.
How high is the tallest spot in the ball’s path?
Answer in units of m.
Firste we need to determine the distance that the ball traveled before it was caught.
The ball path withh form the hypotenuse in a right angle triangle.
Then, let the distance traveled be (h)
We have the speed :
s = 25.1 m/s
Also we have the angle:
a= 24 degrees.
Also, given the distance from the througher to the catcher
But we know that:
cosa = adjacent/ hypotenuse
==> cos 24 = 47.7257 / h
==> h= 47.7257 / cos 24
= 47.7257 / 0.9135
==> h= 52.24 m ( approx.)
Then the ball travelled 52.24 m .
The speed of the ball which is thrown with an angle 24 degree to the horizontal is 25.1 m/s.So the ball has the initial vertical component of velocity 25.1sin24 m/s.
The ball goes on loosing its vertical component of velocity on account of acceleration due to gravity.At the highest point in its path , the ball has the vertical component of velocity 0.
Therefore, we apply the equation of motion v^2 - u^2= 2gh , where v is the final vertical component at the highest point which is zero, u is the initial vertical component of velocity which is 25.1 sin24, g (= 9.81m/s^2) is the acceleration due due to gravity, and h is highest height the ball reaches in its path.
Substituting the values v = 0, u = 25.1sin24 , g = -9.82m/s^2, we determine h.
0^2 - (25.1sin24)^2 = 2*(-9.81)h.
h = (25.1sin24)^2/ (2*9.981) .
h = 5.3122 meter.
Therefore the ball reaches the highest point of 5.3122m in its path.