AS Daily A Level Physics question
A ball is launched straight up from a bench in a lab. Take upward as positive and assume a constant downward acceleration of 9.8 m s−2. A motion sensor shows that in the first 0.50 s after launch, the displacement is +2.0 m. Over the next 0.50 s (from t = 0.50 s to t = 1.00 s), what is the displacement relative to the position at t = 0.50 s?
Answer
The correct answer is A.
Correct: A — ≈ −0.45 m (downward). For equal time steps under constant acceleration, the second interval’s displacement equals the first plus a·(Δt)^2, so s2 = 2.0 m + (−9.8 m s−2)·(0.50 s)^2 = 2.0 − 2.45 = −0.45 m; the negative sign means downward. A — This matches the sign convention and uses the interval relationship s2 = s1 + a(Δt)^2; a quick limiting check (if g → 0, s2 → s1) is also satisfied. B — Sign error: it assumes the net displacement is still upward; however the object passes its peak within this interval so the net displacement is downward, and numerically 2.0 − 2.45 < 0. C — Assumes equal displacements in equal times (as if acceleration were zero); that limiting case would give +2.0 m only if a = 0, which is not true under gravity. D — Assumes increasing upward motion (as if acceleration were upward), contradicting the downward acceleration; doubling the first interval’s displacement is incompatible with deceleration on the way up.