Question

A ball is thrown straight up at 13.0 m/s near the edge of a 30.0-meter high cliff. The ball lands at the bottom of the cliff. a) What is the acceleration of the ball just after it is thrown (magnitude and direction)? b) What is the acceleration of the ball at the peak of its trajectory (magnitude and direction)? c) What is the velocity of the ball just before impact at the bottom of the cliff? d) How much time after being thrown does the ball take to land?

Answer 1

a/b the acceleration of the ball is g (gravity)

the free-fall equation is
... h = -1/2 g t^2 + 13.0 t + 30.0
... h is the height from the bottom of the cliff
... t is the flight time

d) plug in zero for h to find the flight time

c) the time at peak is on the axis of symmetry of the free-fall equation ... t = -13.0 / -g

subtract the time at peak from the flight time to find the fall time

multiply the fall time by g to find the impact velocity (remember direction and magnitude)

Answer 2

a = g = -9.8 m/s^2.

b. a = g = -9.8 m/s^2.

c. V = Vo + g*Tr = 0.
13 - 9.8Tr = 0
Tr = 1.33 s. = Rise time.

h = ho + Vo*Tr + 0.5g*Tr^2.
h = 30 + 13*1.33 - 4.9*1.33^2 = 38.6 m.

V^2 = Vo^2 + 2g*h.
V^2 = 13^2 - 19.6*38.6 = 925.6
V = 30.4 m/s.

d. h = 0.5g*Tf^2 = 38.6.
4.9*Tf^2 = 38.6
Tf = 2.81 s. = Fall time.

Tr + Tf = 1.33 + 2.81 = 4.12 s. = Time in flight.