A ball thrown vertically upwards returns to the hand in 4.8s. (a.) How high did the ball go? (b.) What was the ball's initial speed? (c.) At what time(s) is the ball's speed equal to 19.6 m/s?
To solve this problem, we can use the equations of motion for vertical motion.
(a.) How high did the ball go?
The equation we can use to find the maximum height (h) is:
h = v₀t + (1/2)at²
Since the ball goes up and then returns to the hand, the total time taken would be twice the time it took to reach the maximum height.
So, the total time (2t) is 4.8 seconds.
Let's substitute the values into the equation:
h = 0 + (1/2)gt²
h = (1/2)(9.8 m/s²) * (4.8 s)²
Simplifying the equation:
h = 0.5 * 9.8 * (4.8)²
h ≈ 56.41 meters
Therefore, the ball goes to a height of approximately 56.41 meters.
(b.) What was the ball's initial speed?
The equation we can use to find the initial speed (v₀) is:
v = v₀ + at
At the maximum height, the vertical velocity (v) will be zero. The acceleration (a) is gravity, which is -9.8 m/s².
Let's substitute the values into the equation:
0 = v₀ + (-9.8 m/s²) * (2t)
Simplifying the equation:
0 = v₀ - 19.6t
Solving for v₀:
v₀ = 19.6t
Since we already know the total time is 4.8 seconds, we can substitute the value to calculate the initial speed:
v₀ = 19.6 m/s * 4.8 s
v₀ ≈ 94.08 m/s
Therefore, the ball's initial speed was approximately 94.08 m/s.
(c.) At what time(s) is the ball's speed equal to 19.6 m/s?
To find the time when the ball's speed is equal to 19.6 m/s, we can use the equation:
v = v₀ + at
Let's substitute the values into the equation:
19.6 = 94.08 + (-9.8) * t
Simplifying the equation:
-74.48 = -9.8t
Solving for t:
t = -74.48 / -9.8 ≈ 7.6 seconds
Therefore, the ball's speed is equal to 19.6 m/s at approximately 7.6 seconds.