a ball is thrown upward from a bridge with a speed of 48 ft/s. It misses the bridge on way down and lands in water 160 ft. below. Find how long the ball rises, how high it goes, how long it is in the air, and its velocity when it strikes the water.
just start with the usual functions for height and velocity, and then solve for the values you need
v = 48-16t
s = 160+48t-8t^2
assuming the water is at height 0.
To solve this problem, we can apply the laws of motion to the vertical motion of the ball. We'll use the kinematic equations to find the time it takes for the ball to rise, the maximum height reached, the total time in the air, and the velocity when it strikes the water.
Let's break down the problem step by step:
1. Find the time it takes for the ball to rise:
We know the initial upward velocity is 48 ft/s, and the only force acting on the ball is gravity. The initial velocity will gradually decrease until it reaches zero at the highest point. At that point, the ball starts moving downward. We can use the first equation of motion:
vf = vi + at
Where:
vf = final velocity (0 ft/s at the highest point)
vi = initial velocity (48 ft/s)
a = acceleration due to gravity (-32.2 ft/s^2, since it acts in the opposite direction to the motion)
Rearranging the equation for time (t), we get:
t = (vf - vi) / a
t = (0 - 48) / (-32.2)
t = 1.49 seconds (approximately)
2. Find the maximum height reached by the ball:
To find the maximum height, we can use the second equation of motion:
s = vit + (1/2)at^2
Where:
s = displacement (maximum height)
vi = initial velocity (48 ft/s)
t = time taken to reach maximum height (1.49 seconds, as found in step 1)
a = acceleration due to gravity (-32.2 ft/s^2)
Plugging in the values, we get:
s = (48 * 1.49) + (0.5 * (-32.2) * (1.49)^2
s ≈ 35.86 ft (approximately)
3. Find the total time the ball is in the air:
The total time in the air is twice the time it takes to reach the maximum height:
Total time = 2 * 1.49
Total time ≈ 2.98 seconds (approximately)
4. Find the velocity when the ball strikes the water:
To find the velocity, we can use the third equation of motion:
vf^2 = vi^2 + 2as
Where:
vf = final velocity (unknown)
vi = initial velocity (48 ft/s)
a = acceleration due to gravity (-32.2 ft/s^2)
s = displacement (160 ft)
Plugging in the values, we get:
vf^2 = (48)^2 + 2 * (-32.2) * (-160)
vf ≈ 107.87 ft/s (approximately)
So, to summarize:
- The ball rises for approximately 1.49 seconds.
- The maximum height reached by the ball is approximately 35.86 ft.
- The total time the ball is in the air is approximately 2.98 seconds.
- The velocity when the ball strikes the water is approximately 107.87 ft/s.