Use a(t)=-32 ft/second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with an initial velocity of 96 feet per second. How high will it go?
Well, well, well... looks like this ball is trying to reach for the stars! Let's find out how high it's gonna go with all that ambition.
To solve this, we'll use the good old "kinematic equations." First, we need to find out how long it takes for the ball to reach its peak. We know the initial velocity is 96 feet per second, and the acceleration is -32 ft/s² (downwards due to gravity).
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time, we can find the time it takes to reach its peak.
Since the ball comes to a stop at its peak, we can say that v = 0. So, 0 = 96 + (-32)t. Solving for t, we find t = 3 seconds.
Now, since we know the time it takes to reach the peak, we can find the height it reaches using the equation s = ut + (1/2)at², where s is the height.
So, s = (96)(3) + (1/2)(-32)(3)². Doing some calculations, we get s = 144 - 144 = 0.
Wait a minute... Does this mean the ball doesn't go anywhere?! Oh no, looks like this ball's ambition was a bit misguided. In this case, the ball only reaches a height of zero feet. I guess it didn't quite make it to the stars after all! Keep trying, little ball!
To find out how high the ball will go, we need to determine the maximum height it reaches.
Given:
Acceleration due to gravity, a(t) = -32 ft/second^2
Initial velocity, u = 96 ft/second
Step 1: Determine the time it takes for the ball to reach the highest point.
The velocity of the ball at its highest point will be zero. Therefore, we can use the equation:
v = u + at
Where:
v = final velocity (0 ft/second)
u = initial velocity (96 ft/second)
a = acceleration due to gravity (-32 ft/second^2)
t = time taken
0 = 96 + (-32)t
Rearranging the equation, we have:
32t = 96
Dividing both sides of the equation by 32:
t = 3 seconds
Step 2: Calculate the maximum height reached by the ball.
To determine the maximum height, we need to use the equation:
s = ut + (1/2)at^2
Where:
s = maximum height
u = initial velocity (96 ft/second)
t = time taken (3 seconds)
a = acceleration due to gravity (-32 ft/second^2)
Substituting the values into the equation:
s = (96)(3) + (1/2)(-32)(3)^2
s = 288 + (-16)(9)
s = 288 - 144
s = 144 feet
Therefore, the ball will reach a maximum height of 144 feet.
To find how high the ball will go, we need to determine the maximum height reached by the ball. We can calculate this using the following formula:
h = (v^2 - u^2) / (2a),
where:
- h is the height reached by the ball,
- v is the final velocity of the ball at its maximum height,
- u is the initial velocity of the ball, and
- a is the acceleration due to gravity, which is given as -32 ft/second squared (negative because it acts in the opposite direction to the ball's motion).
Let's substitute the given values into the formula:
h = (v^2 - u^2) / (2a)
= (0 - (96 ft/s)^2) / (2(-32 ft/s^2))
= (-9216 ft^2/s^2) / (-64 ft/s^2)
= 144 ft^2/s^2.
To simplify the units, we can divide the numerator by the denominator:
h = 144 ft^2/s^2 / 1 ft^2/s^2
= 144 ft.
Therefore, the ball will reach a maximum height of 144 feet.
height = -16t^2 + 96t
v = -32t + 96
= 0 at the top
32t=96
t = 3
it will take 3 seconds to reach the max, and that
max is
-16(9) + 96(3) = 144 ft