A student throws a ball
vertically upward from the ground with an initial velocity of 25
m/S.
Questions:
a) How long will it take the ball to reach its maximum height?
b) What is the maximum height it will reach?
c) How long will the ball be in the air before it hits the ground again?
d) What is the velocity of the ball
just before it touches the ground?
Answer (1)
Answer:We assume standard gravity, =9.8 m/s2g=9.8m/s2, and no air resistance.Known:Initial velocity upward: =25 m/su=25m/sAcceleration due to gravity: =9.8 m/s2g=9.8m/s2 (downward)(a) Time to reach maximum heightAt maximum height, final velocity =0v=0.=−v=u−gt0=25−9.80=25−9.8t=259.8t=9.825 ≈2.55 secondst≈2.55seconds(b) Maximum heightUse formula:ℎ=22h=2gu2 ℎ=2522(9.8)=62519.6h=2(9.8)252 =19.6625 ℎ≈31.9 metersh≈31.9meters(c) Total time in airTime up = 2.55 2.55s.Time down is the same (symmetry, ignoring air resistance).=2×2.55≈5.10 secondsT=2×2.55≈5.10seconds(d) Velocity just before it hits the groundBy symmetry, the velocity magnitude equals the initial velocity, but downward.=−25 m/sv=−25m/s(negative sign = downward direction).✅ Final Answers:a) 2.55 s2.55sb) 31.9 m31.9mc) 5.10 s5.10sd) −25 m/s−25m/s (downward)
