A stone is dropped from a height of 49 m and simultaneously another ball is
thrown upwards from the ground with a speed of 40 m/s. When and where do
the two stones meet?
Answers
Answer:
You mentioned this speed of 40m/s because this is the speed that the stone which is thrown from a height of 49m will reach when it touches the ground. which gives: t=49/40=1.225 second.
Explanation:
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Answer:
If ground level is at a height y=0m with the positive direction upwards, then at time t=0s , we have Stone 1 on the ground with y1,0=0m and v1,0=40ms , and Stone 2 at a height of y2,0=49m and v2,0=0ms . Since the acceleration due to gravity is constant, we can use the equation y=y0+v0t+12at2 for both stones, setting y1,t for Stone 1 equal to y2,t for Stone 2 to find the time, t , at which the height, y , is the same for both:
y1,t=0m+(40ms)t+12(−9.81ms2)t2
y2,t=49m+(0ms)t+12(−9.81ms2)t2
y1,t=y2,t
(40ms)t=49m
t=1.225s
Substituting this value for t into either equation for y , we find that the two stones meet at a height of y=41.639m above the ground.
Explanation: