A ball is thrown from the edge of a cliff with an initial velocity of 60·m/s upward. Answer the following questions using + (upward) and - (downward) to indicate the direction of the velocity. Ignore air resistance and assume the ball does not hit the ground during the question.

Question

(a) Find the velocity of the ball 5 seconds after it is thrown. 1 m/s.

(b) Find the velocity of the ball 10 seconds after it is thrown. 2 m/s.

(c) How high does it rise above the edge of the cliff (from where it was thrown)? 3 m.

Answer 1

call g = -10m/s^2

v = vi - g t = +60 - 10t
when t = 5
v = 60 -10(5) = 10 m/s

when t = 10
v = 60 - 10 (10) = -40 m/s

y = yi + vi t - (1/2)g t^2
what is t at the top where v = 0 ??
0 = 60 - 10 t
t = 6 seconds to top
y = 0 + 60(6) -5 (36)
= 180