Question

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

Answer 1

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