A ball is thrown vertically upward from ground level with initial velocity of 96 feet per second. Assume the acceleration of the ball is a(t) = -32 ft^2 per second. (Neglect air Resistance.)
(a) How long will it take the ball to raise to its maximum height? What is the maximum heights?
(b) After how many seconds is the velocity of the ball one-half the initial velocity?
(c) What is the height of the ball when its velocity is one-half the initial velocity?
(a) Acceleration is Change in velocity over time, thus we can write it as,
a = (vf - vo) / t
Note that final velocity (vf) is Zero when it reaches its Maximum height since it stops. And since acceleration and Initial velocity (vo) are given, we can Substitute:
-32 = (0 - 96) / t
-32t = -96
t = 3 seconds
For Maximum height, we can use the formula,
h = vo*t - (1/2)(a)(t^2)
h = 96*3 - (1/2)(-32)(3^2)
h = 144 ft
(c) We can calculate for the height of the ball when it is 1/2 of Initial velocity using the formula,
vf^2 - vo^2 = 2gd
Since vf= 1/2*vo,
(1/2*vo)^2 - vo^2 = 2gd
Substituting,
(1/2*96)^2 - 96^2 = 2*(-32)*d
48^2 - 96^2 = -64d
d = -6912/-64
d = 108 ft
(b) We can calculate the time when the ball is moving at 1/2 of its Initial velocity with the same formula we used in (a),
h = vo*t - (1/2)(a)(t^2)
Substitute h = 108, and solve for t.
(a) How long will it take the ball to raise to its maximum height? What is the maximum height?
Well, let's throw some humor into the mix, shall we? It's a bird, it's a plane, no wait, it's a ball! And it's going up, up, and away!
To determine the time it takes for the ball to reach its maximum height, we need to find the time when its velocity becomes zero.
Using the equation for vertical motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can solve for t.
Since the ball is going upwards, the final velocity will be 0. So, we can plug in the values:
0 = 96 - 32t
Now, let's solve for t. Dividing both sides by -32, we get t = 3.
So, it will take the ball 3 seconds to reach its maximum height.
To find the maximum height, we can use the equation for displacement, s = ut + 0.5at^2, where s is the displacement or height.
Plugging in the values, we get s = 96(3) + 0.5(-32)(3)^2.
Calculating this out, we find that the maximum height is 144 feet. That's a pretty high leap for a ball, don't you think?
(b) After how many seconds is the velocity of the ball one-half the initial velocity?
Well, let's put our detective hats on and solve this mystery!
We know that the initial velocity of the ball is 96 feet per second, so we need to find the time when the velocity is one-half of that, or 48 feet per second.
Using the equation v = u + at, we can plug in the values:
48 = 96 - 32t
Let's solve this equation. Substracting 96 from both sides, we get -48 = -32t.
Dividing both sides by -32, we find t = 1.5.
So, after 1.5 seconds, the velocity of the ball is one-half of the initial velocity. That's one speedy ball!
(c) What is the height of the ball when its velocity is one-half the initial velocity?
Ah, the slow and steady race for the height!
When the velocity is one-half the initial velocity, we know the velocity is 48 feet per second. Now, we need to find the corresponding height.
Using the equation s = ut + 0.5at^2, we can plug in the values:
s = 96(1.5) + 0.5(-32)(1.5)^2
Calculating this out, we find that the height of the ball when its velocity is one-half the initial velocity is 54 feet. Well, the ball may be slower, but it's still reaching for the stars (or quite high up in the air)!
Keep those questions coming, and I'll keep the humor flowing!
To solve these problems, we can use the equations of motion for constant acceleration.
Let's denote:
Vi = initial velocity = 96 ft/s
a = acceleration = -32 ft/s^2
t = time (unknown)
Vf = final velocity (unknown)
H = height (unknown)
(a) To find the time it takes the ball to reach its maximum height, we can use the equation:
Vf = Vi + at
At the maximum height, the final velocity will be 0 ft/s. Plugging in the known values:
0 = 96 - 32t
Solving for t:
32t = 96
t = 96 / 32
t = 3 seconds
So, it will take the ball 3 seconds to reach its maximum height.
To find the maximum height, we can use the equation:
H = Vi * t + (1/2) * a * t^2
Plugging in the known values:
H = 96 * 3 + (1/2) * (-32) * (3^2)
H = 288 + (-48)
H = 240 ft
Therefore, the maximum height the ball reaches is 240 feet.
(b) To find the time when the velocity of the ball is one-half the initial velocity, we can solve for t in the equation:
Vf = Vi + at
Plugging in the known values:
Vf = 96/2 = 48 ft/s
48 = 96 - 32t
Solving for t:
32t = 96 - 48
t = 48 / 32
t = 1.5 seconds
So, after 1.5 seconds, the velocity of the ball is one-half the initial velocity.
(c) To find the height of the ball when the velocity is one-half the initial velocity, we can use the equation:
H = Vi * t + (1/2) * a * t^2
Plugging in the known values:
H = 96 * 1.5 + (1/2) * (-32) * (1.5^2)
H = 144 - 36
H = 108 ft
Therefore, the height of the ball when its velocity is half the initial velocity is 108 feet.
To answer these questions, we can apply the principles of kinematics. Kinematics is the study of how objects move without considering why they move.
(a) How long will it take the ball to reach its maximum height, and what is the maximum height?
To solve for the time it takes the ball to reach its maximum height, we need to find the time at which the vertical velocity becomes zero. We can use the following equation of motion:
v = u + at
where:
v = final velocity (in this case, 0 ft/s since the ball reaches its maximum height)
u = initial velocity (96 ft/s)
a = acceleration (-32 ft/s^2, negative because it acts in the opposite direction of motion)
t = time
Rearranging the equation, we have:
t = (v - u) / a
Plugging the values into the equation, we get:
t = (0 - 96) / -32 = 3 seconds
So, it takes 3 seconds for the ball to reach its maximum height.
To find the maximum height, we can use the equation:
s = ut + (1/2)at^2
where s is the displacement (maximum height) and all the other variables have the same meaning as before.
Since we want to find the displacement at the maximum height, the equation simplifies to:
s = ut + (1/2)at^2 = 96 * 3 + (1/2) * (-32) * (3^2) = 144 ft
Therefore, the maximum height of the ball is 144 feet.
(b) After how many seconds is the velocity of the ball one-half the initial velocity?
The velocity of the ball at any given time can be found using the equation we used before:
v = u + at
We need to find the time when the velocity, v, is half the initial velocity, u.
Setting v = (1/2)u, we have:
(1/2)u = u - 32t
Now we can solve for t:
(1/2)u = u - 32t
(1/2)(96) = 96 - 32t
48 = 96 - 32t
32t = 48
t = 48 / 32 = 1.5 seconds
Therefore, the velocity of the ball is half the initial velocity after 1.5 seconds.
(c) What is the height of the ball when its velocity is one-half the initial velocity?
To find the height of the ball at this time, we need to use the equation we used before:
s = ut + (1/2)at^2
Substituting the values, we have:
s = 96 * 1.5 + (1/2) * (-32) * (1.5^2)
s = 144 - 36 = 108 ft
Thus, the height of the ball when its velocity is one-half the initial velocity is 108 feet.