A physics student throws a softball straight up into the air. The ball was in the air for a total of 6.16 s before it was caught at its original position.
(a) What was the initial velocity of the ball?
(b) How high did it rise?
Tr + Tf = 6.16 s,
Tf = Tr,
Tr + Tr = 6.16,
2Tr = 6.16,
Tr = 3.08 s. = Rise time or time to reach max. heightt.
a. Vf = Vo + gt,
Vo = Vf - gt,
Vo = 0 - (-9.8)*3.08 = 30.2 m/s.
b. h = Vo*t + 0.5g*t^2,
h = 30.2*3.08 - 4.9*(3.08)^2 = 46.53 m.
To solve this problem, we can use the equations of motion for an object in free fall. We'll assume the acceleration due to gravity is -9.8 m/s^2 since the ball is thrown straight up and comes back down.
(a) To find the initial velocity of the ball, we can use the equation:
v = u + at
where:
v is the final velocity (which is zero at the top of the ball's path),
u is the initial velocity,
a is the acceleration (-9.8 m/s^2), and
t is the time taken to reach the top of the ball's path.
Since the final velocity is zero, the equation becomes:
0 = u - 9.8 * t
We are given that the ball was in the air for a total of 6.16 s. So we can substitute this value into the equation:
0 = u - 9.8 * 6.16
Simplifying the equation, we get:
u = 9.8 * 6.16
Calculating this, we find:
u ≈ 60.128 m/s (rounded to three decimal places)
Therefore, the initial velocity of the ball was approximately 60.128 m/s.
(b) To find the height the ball reached, we can use the equation:
s = ut + (1/2)at^2
where:
s is the height,
u is the initial velocity (which we just found to be 60.128 m/s),
t is the total time in the air (6.16 s), and
a is the acceleration (-9.8 m/s^2).
Substituting the values into the equation:
s = 60.128 * 6.16 + (1/2) * (-9.8) * (6.16)^2
Simplifying the equation, we get:
s = 369.6512 + (-198.7696)
Calculating this, we find:
s ≈ 170.8816 m (rounded to three decimal places)
Therefore, the ball reached a maximum height of approximately 170.882 m.
To answer these questions, we will use the following equations of motion:
1. The formula for calculating the height reached by an object thrown vertically upward is given by:
h = (v₀² - v²) / (2 * g),
where:
- h is the maximum height reached,
- v₀ is the initial velocity, and
- v is the final velocity, which is 0 when the ball reaches its highest point.
2. The formula for determining the final velocity of an object thrown vertically upward is given by:
v = v₀ - g * t,
where:
- v is the final velocity,
- v₀ is the initial velocity,
- g is the acceleration due to gravity (approximately 9.8 m/s²), and
- t is the time in seconds.
Now, let's solve each part of the problem step by step:
(a) What was the initial velocity of the ball?
To find the initial velocity (v₀), we need the final velocity (v) and the time (t).
Given:
- v = 0 m/s (since the ball stops at its highest point),
- t = 6.16 s.
Using the formula for final velocity, we have:
0 = v₀ - (9.8 m/s²) * 6.16 s.
Rearranging the equation to solve for v₀, we get:
v₀ = (9.8 m/s²) * 6.16 s.
Evaluating this expression, we find:
v₀ ≈ 60.13 m/s.
Therefore, the initial velocity of the ball was approximately 60.13 m/s.
(b) How high did it rise?
Using the formula for maximum height reached, we can calculate the height:
h = (v₀² - v²) / (2 * g).
Given:
- v₀ ≈ 60.13 m/s (from part (a)),
- v = 0 m/s (since the ball stops at its highest point), and
- g = 9.8 m/s².
Plugging in these values, we have:
h = (60.13 m/s)² / (2 * 9.8 m/s²).
Calculating this expression, we find:
h ≈ 185.71 m.
Therefore, the ball reached a maximum height of approximately 185.71 meters.