Asked by Kimberly

A ball is thrown horizontally from the top of a building 43.4 m high. The ball strikes the ground at a point 57.3 m from the base of the building.
The acceleration of gravity is 9.8 m/s^2.
a. Find the time the ball is in motion.
Answer in units of s.
b. Find the initial velocity of the ball. Answer in units of m/s.
c. Find the x component of its velocity just before it strikes the ground. Answer in units of m/s.
d. Find the y component of its velocity just before it strikes the ground. Answer in units of m/s.


Answered by Henry
a. h = 0.5g*t^2
h = 43.4 m.
g = 9.8 m/s^2
t = Fall time.
Solve for t.

b. Dx = Xo * t
Dx = 57.3 m. = Hor. dist.
Xo = Initial hor velocity.
t = Fall time calculated in part a.
Solve for Xo.

c. X = Xo=Initial hor velocity(part b).
Xo does not change.

d. Y = Yo + g*t
Yo = 0
g = 9.8 m/s^2
t = Fall time calculated in part a.
Solve for Y.


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