A ball is tossed straight upward. Take upward as the positive direction. (a) What is the sign of its displacement during the ascent, from the release point to the highest point? (b) What is the sign of its displacement during the descent, from the highest point back to the release point? (c) What is the ball’s acceleration at its highest point? Take g = 9.8 m/s².
Explanation
(a) During ascent, the ball moves upward from its release point.
Since upward is defined as positive, its displacement is positive.
(b) During descent, the ball moves from the highest point back down to the release point.
Its final position is below its initial position for this part of the motion, so its displacement is negative.
(c) Gravity acts downward throughout the motion, including at the highest point.
The velocity is momentarily zero at the highest point, but acceleration is not zero.
A pitcher tosses a baseball vertically upward along the y-axis with an initial speed of 12 m/s. (a) How long does the ball take to reach its maximum height? (b) What is its maximum height above the release point? (c) How long does it take to reach a point 5 m above its release point? Take g = 9.8 m/s² and neglect air resistance.
Explanation
(a) At maximum height, the vertical velocity is zero.
Using the equation of motion v = u − gt:
0 = u − gt.
t_max = u/g = 12/9.8 = 1.22 s.
(b) Use v² = u² − 2gH at maximum height, where v = 0.
H_max = u²/(2g).
H_max = (12)²/(2 × 9.8) = 7.35 m.
(c) Use the equation of motion y = ut − ½gt².
5 = 12t − ½ × 9.8 × t².
Rearranging: 4.9t² − 12t + 5 = 0.
Using the quadratic formula gives two positive times:
t₁ = (u − √(u² − 2gh))/g = 0.53 s.
t₂ = (u + √(u² − 2gh))/g = 1.92 s.
The first time occurs during ascent and the second during descent.
✓
Conclusion
(a) Time to maximum height = 1.22 s. (b) Maximum height = 7.35 m. (c) The ball reaches 5.0 m above its release point at 0.53 s on the way up and 1.92 s on the way down.