A ball is thrown vertically upward from the ground with an initial velocity of 103
ft per sec.
Determine what the maximum height will be
it stops at the top, so the average velocity is ... (103 + 0) / 2
gravitational acceleration is
32 ft/s^2 , downward
so the time to reach the peak is
... 103 ft/s / 32 ft/s^2
r * t = d
To determine the maximum height reached by the ball, we can use the equation of motion in vertical direction:
h = (v^2 - u^2) / (2g)
where:
h is the maximum height,
v is the final velocity (0 ft/sec at maximum height),
u is the initial velocity (103 ft/sec),
g is the acceleration due to gravity (32.2 ft/sec^2).
Let's plug in the values and calculate the maximum height:
h = (0^2 - 103^2) / (2 * 32.2)
= (-10609) / 64.4
≈ -164.84 ft
Since the maximum height cannot be negative, we discard the negative sign. Therefore, the maximum height reached by the ball is approximately 164.84 ft.
To determine the maximum height reached by the ball, we can use the equations of motion.
First, we need to find the time taken by the ball to reach its maximum height. We can use the equation:
vf = vi + at
Where:
vf is the final velocity (which is 0 at the maximum height since the ball momentarily stops moving before falling back down)
vi is the initial velocity (103 ft/sec)
a is the acceleration due to gravity (-32 ft/sec^2, considering upward as positive and downward as negative)
t is the time taken
Plugging in the values, we have:
0 = 103 - 32t
Solving for t:
32t = 103
t = 103 / 32
t ≈ 3.219 seconds
Now, we can find the maximum height by using the equation:
h = vi*t + (1/2)*a*t^2
Where:
h is the maximum height
vi is the initial velocity (103 ft/sec)
a is the acceleration due to gravity (-32 ft/sec^2)
t is the time taken (3.219 seconds)
Plugging in the values, we have:
h = 103*3.219 + (1/2)*(-32)*(3.219)^2
Calculating:
h ≈ 331.99 feet
Therefore, the maximum height reached by the ball is approximately 331.99 feet.