1st Ball:
d1 = Vo*t + 0.5g*t^2
d1 = 0 + 4.9*2^2 = 19.6 m in 2 seconds
2nd Ball:
d2 = d1 + 19.6 m
Vo*t + 0.5g*t^2 = 4.9t^2 + 19.6
30t + 4.9t^2 - 4.9t^2 = 19.6
30t = 19.6
t = 0.6533 s. To overtake the 1st ball.
A ball is dropped a high cliff, and two second later another ball was thrown vertically downwards with an initial speed of 30 m/s.How long will it take the second ball to overtake the first?
3 answers
Correction:
1st Ball:
d1 = Vo*t + 0.5g*t^2
d1 = 0 + 4.9*2^2 = 19.6 m. in 2 s.
V = Vo + g*t = 0 + 9.8*2 = 19.6 m/s.
2nd Ball:
d2 = d + 19.6 m.
Vo*t + 0.5g*t^2 = V*t + 0.5g*t^2 + 19.6
Vo*t - V*t + 0.5g*t^2-0.5g*t^2 = 19.6
Vo*t - V*t = 19.6
30t- 19.6t = 19.6
10.4t = 19.6
t = 1.88 s. To overtake the 1st ball.
1st Ball:
d1 = Vo*t + 0.5g*t^2
d1 = 0 + 4.9*2^2 = 19.6 m. in 2 s.
V = Vo + g*t = 0 + 9.8*2 = 19.6 m/s.
2nd Ball:
d2 = d + 19.6 m.
Vo*t + 0.5g*t^2 = V*t + 0.5g*t^2 + 19.6
Vo*t - V*t + 0.5g*t^2-0.5g*t^2 = 19.6
Vo*t - V*t = 19.6
30t- 19.6t = 19.6
10.4t = 19.6
t = 1.88 s. To overtake the 1st ball.
how come the answer they got was t=3.88s