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.
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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.
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