An inquisitive physics student and mountain climber climbs a 51.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.02 m/s.
(a) How long after release of the first stone do the two stones
hit the water?
(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?
|direction||---Select--- upward downward|
(c) What is the speed of each stone at the instant the two stones hit the water?
Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.
a) Vo*t + 0.5g*t^2 = d.
1.92*t + 4.9*t^2 = 49.
4.9t^2 + 1.92t - 49 = 0.
Use Quadratic Formula and get:
t = 2.9714 seconds.
b. Vo*2.9714 + 4.9*2.9714^2 = 49.
Vo*2.9714 + 43.263 = 49.
2.9714*Vo = 49 - 43.263 = 5.7368.
Vo = 5.7368 / 2.9714 = 1.93068 m/s. ----> Intial Velocity
Direction Vertically Downward
c. Vf^2 = Vo^2 + 2g*d.
Vf^2 = 1.92^2 + 19.6*49 = 964.086.
Vf = 31.049 m/s. = Final velocity.
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