A 0.175-lb ball is thrown upward with an initial velocity of 75.0 ft/s. What is the
maximum height reached by the ball?
Would I need to set it up like
√2*75.0*-32.2/0/175?
V^2 = Vo^2 + 2g*h = 0 @ max ht.
h = -(Vo^2)/2g = -(75^2)/-64 = 87.9 Ft.
To find the maximum height reached by the ball, you can use the equation of motion for vertically thrown objects. The equation is:
h = (v^2) / (2g)
Where:
h is the maximum height
v is the initial velocity
g is the acceleration due to gravity (approximately -32.2 ft/s^2)
Now, substitute the given values into the equation:
h = (75.0 ft/s)^2 / (2 * -32.2 ft/s^2)
Calculating this expression step by step:
h = (5625 ft^2/s^2) / (-64.4 ft/s^2)
h ≈ -87.244 ft
Since height cannot be negative in this context, we take the positive value:
The maximum height reached by the ball is approximately 87.244 feet.
To find the maximum height reached by the ball, you need to use the equations of motion and apply them to the given situation.
First, let's consider the equation for the vertical motion of the ball:
h = h0 + v0t - (1/2)gt^2
Where:
- h is the final height (maximum height reached),
- h0 is the initial height (which we assume to be 0 in this case),
- v0 is the initial velocity (75.0 ft/s),
- g is the acceleration due to gravity (-32.2 ft/s^2), and
- t is the time taken.
Since we are interested in the maximum height, we need to find the time it takes for the ball to reach its peak. At this point, the vertical velocity (v) will be 0. We know that the initial velocity (v0) is 75.0 ft/s, and we assume the final velocity (vf) is 0 ft/s.
Using the equation for vertical velocity:
vf = v0 + gt
We can solve for the time (t):
0 = 75.0 - 32.2t
Simplifying and solving for t:
32.2t = 75.0
t = 75.0 / 32.2
t ≈ 2.33 seconds.
Now that we have the time it takes for the ball to reach its peak, we can substitute this value into the equation for height:
h = h0 + v0t - (1/2)gt^2
Plugging in the values:
h = 0 + 75.0 * 2.33 - (1/2) * (-32.2) * (2.33)^2
Simplifying:
h ≈ 86.4 feet
Therefore, the maximum height reached by the ball is approximately 86.4 feet.