Find the maximum height reached by a ball thrown upward at a velocity of 95ft/s
H = Vo^2/(2*g)
Use g = 32.2 ft/s^2
H = 140 feet
To find the maximum height reached by a ball thrown upward at a velocity of 95 ft/s, we can use the kinematic equation for vertical motion:
h = (v^2) / (2g)
Where:
- h is the maximum height
- v is the initial velocity
- g is the acceleration due to gravity (approximately equal to 32.2 ft/s^2).
Plugging in the values, we get:
h = (95 ft/s)^2 / (2 * 32.2 ft/s^2)
Simplifying further:
h = 9025 ft^2/s^2 / 64.4 ft/s^2
h ≈ 140.09 ft
Therefore, the maximum height reached by the ball is approximately 140.09 ft.
To find the maximum height reached by a ball thrown upward at a velocity of 95 ft/s, we can use the equations of motion and the concept of projectile motion.
The key equation we'll need is the vertical displacement equation for an object in projectile motion:
y = v0t - (1/2)gt^2
where:
y = vertical displacement (height)
v0 = initial velocity in the vertical direction (upward)
t = time
g = acceleration due to gravity (approximately 32 ft/s^2)
In this case, the initial velocity (v0) is 95 ft/s, and we want to find the maximum height (y). To find the time it takes for the ball to reach its maximum height, we need to determine when the velocity becomes zero. At the maximum height, the velocity is zero because the ball stops momentarily before falling back down.
Using the equation for velocity in the vertical direction:
v = v0 - gt
At the maximum height, v = 0:
0 = 95 - 32t
32t = 95
t = 95 / 32
t ≈ 2.97 seconds (approximately)
Now that we have the time it takes to reach the maximum height, we can substitute this value back into the vertical displacement equation to find the maximum height (y):
y = v0t - (1/2)gt^2
y = (95 ft/s)(2.97 s) - (1/2)(32 ft/s^2)(2.97 s)^2
y ≈ 140.9 ft
Therefore, the maximum height reached by the ball thrown upward at a velocity of 95 ft/s is approximately 140.9 ft.