A boy throws a ball up into the air with a speed of 12.3 m/s. The ball has a mass of 0.25 kg. How much gravitational potential energy will the ball have at the top of its flight? (Assume there is no air resistance.)
18.9 J
18.9 J !!!! #APEX
To calculate the gravitational potential energy (GPE) of an object, you need to know its mass, height, and the acceleration due to gravity.
In this case, we know the mass (0.25 kg) and there is no mention of the height. However, since the question asks specifically about the top of the ball's flight, we can assume that the ball has reached its maximum height. At the top of its flight, the ball comes to a temporary stop before falling back down.
When the ball reaches its maximum height, its velocity will be momentarily zero. This means that the initial velocity (12.3 m/s) will change to zero as it reaches the top. Moreover, since the ball is in free fall, the acceleration due to gravity (g) will act to pull it back down towards the ground. The gravitational acceleration on Earth is approximately 9.8 m/s^2.
To calculate the height (h) of the ball at the top of its flight, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (which is zero at the top)
u = initial velocity
a = acceleration
s = distance
In this case, we know u = 12.3 m/s, v = 0 m/s, and a = -9.8 m/s^2 (as the acceleration is acting against the motion). We can rearrange the equation to solve for s:
0^2 = 12.3^2 + 2(-9.8)s
Simplifying the equation gives:
0 = 151.29 - 19.6s
19.6s = 151.29
s = 7.72 m
So, the height of the ball at the top of its flight is 7.72 meters.
Now, we can calculate the gravitational potential energy using the formula:
GPE = mgh
where:
m = mass (0.25 kg)
g = gravity (9.8 m/s^2)
h = height (7.72 m)
Plugging in the values:
GPE = 0.25 kg * 9.8 m/s^2 *7.72 m
Calculating the value gives:
GPE = 1.88 J
Therefore, the ball will have approximately 1.88 Joules of gravitational potential energy at the top of its flight.
same as the KE at the start:
1/2 * .25 * 12.3^2