Respuesta :
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
[tex]x=ut+\frac{1}{2} at^2\\\\x= 0 \times t+\frac{1}{2} gt^2 =\frac{1}{2} gt^2......(1)[/tex]
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)
[tex]y=24t-\frac{1}{2} gt^2.............(2)[/tex]
we know that total distance traveled by ball and stone is 50 m
[tex]x+y=50 m[/tex]
adding equation 1 and 2, we get time t
[tex]x+y=\frac{1}{2} gt^2+24t-\frac{1}{2}gt^2\\50=24t\\t=2.083 s[/tex]
substitute this time in equation 2, we can get the required distance where they collide
[tex]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[/tex]
28.73 m from the base of the cliff collide happen
