Question

A boy on a cliff 49 m high drops a stone. One second later he throws a second stone after the first. They both hit the ground at the same time. With what speed did he throw the second stone?​

Answer 1

Answer:

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1st Stone:

4.9t^2 = 49

t^2 = 10

Tf = 3.16 s. = Fall time.

2nd stone:

h = Vo*t + 0.5g*t^2 = 49

t = 3.16-1 = 2.16 s. = Fall time.

Vo*2.16 + 4.9*2.16^2 = 49

2.16Vo + 22.86 = 49

2.16Vo = 49-22.86 = 26.14

Vo = 12.1 m/s. = Initial velocity.

Answer 2

Answer:

first stone

s=ut +1/2 gt²

49=1/2*9.8*t²

98/9.8=t²

t=√10 seconds

second stone

t=√10-1 seconds

t=u(√10-1)+1/2*9.8*(√10-1)²

√10-1= 2.16

49= u*2.16+1/2 *9.8 *(2.16)²

49=u*2.16+22.8614

26.1385=u*2.16

u=12.101m/s

hope it will help u