A ball dropped from rest, covers three-quarters of the distance to the ground in the last second of its fall.?

Question

(a) From what height was the ball dropped?

(b) What was the total time of the fall?

Hint: Set-up an equation of motion y(t) for the ball, and evaluate it at the two different times implied by the question.

One-body but information about two times.

CAN ANYONE PLEASE HELP!!!

Answer 1

t - total time of the fall

h - from that height was the ball dropped

g = 9.81 m/s² - gravitational acceleration

Equation for the first quarter of the fall is

0.25 h = 0.5 g  (t – 1)²

Equation for the total fall is

h = 0.5 g t²



Solution of the above system of equations is

t = 2 seconds

h = 19.62 meters

Is it clear or should I write more details?

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Answer 2

D = 0.5 AT^2

A=9.8 M/s^2

.75D = 0.5T^2 where T=1  ===>T^2 = 1

===>

.75D = 9.8M/s^2 * 1s^2  The s^2 cancel

D= (9.8m)/0.75

D= 13.07 m

D=0.5 * 9.8 *T^2

13.07 = 0.5 * 9.8 * T^2

4.9T^2 = 13.07

T^2 = 2.67 s

T= 1.16 s