A baseball is thrown at an angle of 24◦ relative to the ground at a speed of 25.1 m/s. The ballis caught 47.7257 m from the thrower.How high is...

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.