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defon
3 years ago
7

ball dropped from top of 50m high cliff stone thrown straight up from bottom with speed of 24. they collide. how far above base

of cliff does this happen
Physics
1 answer:
Tresset [83]3 years ago
7 0

Answer:

28.73 m from the base of the cliff collide happen

Explanation:

Equation for ball dropped from 50 cliff is reaches distance x in t seconds isi given by

x=ut+\frac{1}{2} at^2\\\\x= 0 \times t+\frac{1}{2} gt^2 =\frac{1}{2} gt^2......(1)

stone is thrown up from the bottom with speed u=24 m/s . it reaches distance y when stone collide with ball.(g is negative here)

y=24t-\frac{1}{2} gt^2.............(2)

we know that total distance traveled by ball and stone is 50 m

x+y=50 m

adding equation 1 and 2, we get time t

x+y=\frac{1}{2} gt^2+24t-\frac{1}{2}gt^2\\50=24t\\t=2.083 s

substitute this time in equation 2, we can get the required distance where they collide

y=24t-\frac{1}{2} gt^2\\y=24\times 2.083-\frac{1}{2}\times  9.8\times 2.083^2\\y=28.73 m

28.73 m from the base of the cliff collide happen

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