Asked by Sam

A home run is hit such a way that the baseball just clears a wall 26 m high located 114 m from home plate. The ball is hit at an angle of 35â—¦ to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 2 m above the ground.

What is the initial speed of the ball? The acceleration of gravity is 9.8 m/s2 .
Answer in units of m/s.

How much time does it take for the ball to reach the wall?
Answer in units of s.

Find the speed of the ball when it reaches the wall.

Answer in units of m/s.
Answered by Sam
35 degrees*
Answered by Henry
a. Range = Vo^2*sin(2A)/g.
114 = Vo^2*sin(70)/9.8,
Vo = 34.5 m/s.

b. Xo = Vo*cos35 = 34.5*cos35 = 28.2 m/s. = Hor. component.
Yo = 34.5*sin35 = 19.8 m/s = Ver. component.

Y = Yo + g*Tr.
0 = 19.8 -9.8Tr,
Tr = 2.02s. = Rise time. = Time to reach the wall.

h = ho + Yo*Tr + 0.5g*Tr^2.
h = 2 + 19.8*2.02 - 4.9*2.02^2 = 22m above gnd. = max ht.
If the max. ht. is 22 m, the wall can't
be 26 m high. Check the given ht.

c. Assuming the ball is at max. ht. when it reaches the wall:
V = Xo + Yi = 28.2 + 0i = 28.2 m/s.