Asked by jordan
A ball is thrown vertically upward with an initial velocity of 15.4 m/s. What distance will it have traveled after 2.1 seconds
All Answers 1
Answered by
GPT-5 mini
AI
Take upward positive and g = 9.8 m/s^2.
Time to apex: t_apex = v0/g = 15.4/9.8 ≈ 1.57 s < 2.1 s, so the ball has gone up then partway down.
Maximum height: h_max = v0^2/(2g) = 15.4^2/(2·9.8) ≈ 12.10 m.
Height at t = 2.1 s: y = v0 t − 1/2 g t^2 = 15.4·2.1 − 4.9·(2.1)^2 = 32.34 − 21.609 = 10.731 m.
Total distance traveled = up to the apex + down from apex to current position
= h_max + (h_max − y) = 2 h_max − y ≈ 2·12.104 − 10.731 ≈ 13.48 m.
So the ball has traveled about 13.5 m after 2.1 s.
Time to apex: t_apex = v0/g = 15.4/9.8 ≈ 1.57 s < 2.1 s, so the ball has gone up then partway down.
Maximum height: h_max = v0^2/(2g) = 15.4^2/(2·9.8) ≈ 12.10 m.
Height at t = 2.1 s: y = v0 t − 1/2 g t^2 = 15.4·2.1 − 4.9·(2.1)^2 = 32.34 − 21.609 = 10.731 m.
Total distance traveled = up to the apex + down from apex to current position
= h_max + (h_max − y) = 2 h_max − y ≈ 2·12.104 − 10.731 ≈ 13.48 m.
So the ball has traveled about 13.5 m after 2.1 s.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.