A ball is shot straight up into the air with initial velocity of 47 ft/sec. Assuming that the air resistance can be ignored, how high does it go?
A ball is shot straight up into the air with initial velocity of 47 ft/sec
Free fall, s in ft, t in sec
s = s0 + v0t - 16t^2
height, s0 = 0
velocity, v0 = 47
s = 0 + 47t - 16t^2
s = 47t - 16t^2
Instantaneous velocity, v
v = ds/dt = v0 - 32t
v = 47 ft/sec
v = 47 - 32t
To find time, t
At maximum value or turning point, v = 0
v = 47 - 32t
0 = 47 - 32t
32t = 47
t = 47/32 = 1.47 sec to reach max height
s = 47t - 16t^2
t = 1.47 sec
s = 47(1.47) - 16(1.47)^2
s = 69.09 - 16(2.16)
s = 69.09 - 34.56
s = 34.53 feet
So, it takes 1.47 sec to reach its maximum height of 34.53 feet
Check my math
Well, let's see. The ball is shot straight up into the air with an initial velocity of 47 ft/sec. Ignoring air resistance, it will keep going up until gravity decides to intervene. And gravity is a real party pooper, I tell ya. It pulls things back down to Earth.
Now, we know that the initial velocity is 47 ft/sec, but we need to know how long it takes for gravity to bring the ball to a stop at the top of its trajectory. Once it reaches its peak, the velocity will be zero before it starts coming back down.
Hang on, I’m doing some calculations here. Cue the thinking music…
Ah! Alright, so let's use a little physics magic called kinematic equations. We have the initial velocity, the final velocity (which is zero at the peak), and we're looking for the displacement, which is the height it reaches. I won't bore you with all the details, but according to my calculations, the ball reaches a height of roughly 110.9 feet.
So there you have it. That ball reaches for the stars and makes it to approximately 110.9 feet above the ground before gravity calls it back down. Just remember, gravity always brings things back down to reality. It's a tough job, but someone's gotta do it!
To find the maximum height reached by the ball, we can use the kinematic equation for vertical motion:
h = (v^2 - u^2) / (2g)
where:
h = maximum height reached by the ball
v = final velocity of the ball (which will be 0 at the maximum height)
u = initial velocity of the ball
g = acceleration due to gravity (approximately 32 ft/sec^2)
Given:
u = 47 ft/sec
g = 32 ft/sec^2
Let's plug these values into the equation to find the maximum height:
h = (0^2 - 47^2) / (2 * 32)
h = (-47^2) / 64
h = 2209 / 64
h ≈ 34.52 feet
Therefore, the ball reaches a maximum height of approximately 34.52 feet.
To determine the height the ball reaches, we can use the laws of motion and apply the principles of kinematics.
First, we need to understand that the motion of the ball can be broken down into two phases - the upward phase and the downward phase. During the upward phase, the ball is moving against the force of gravity, causing it to slow down until it reaches its highest point. During the downward phase, the ball is accelerating towards the ground due to gravity.
In the upward phase, the ball's velocity decreases at a rate of 32 ft/sec^2 (the acceleration due to gravity). Using this information, we can find the time it takes for the ball to reach its highest point.
The equation that relates velocity, acceleration, and time is:
v = u + at
where:
v = final velocity (which is 0 when the ball reaches its highest point)
u = initial velocity (47 ft/sec in this case)
a = acceleration due to gravity (-32 ft/sec^2, negative because it opposes the upward motion)
t = time
Using this equation, we can rearrange it to solve for time:
t = (v - u) / a
Substituting the values, we have:
t = (0 - 47) / -32
t = 47 / 32
t ≈ 1.47 seconds
So, it takes approximately 1.47 seconds for the ball to reach its highest point.
To find the height the ball reaches, we can use another equation of motion:
s = ut + (1/2)at^2
where:
s = displacement or height
u = initial velocity
t = time
a = acceleration due to gravity
Substituting the values, we have:
s = (47)(1.47) + (1/2)(-32)(1.47)^2
s = 68.9 - 34.45
s ≈ 34.45 feet
Therefore, the ball reaches a height of approximately 34.45 feet.