A ball thrown up has a height f(t)=250t-25t^2 after t seconds.
(1) find the maximum height
(2) find the velocity when it hits the ground.
The units of measurements have not been supplied, so they will not be used. On the other hand, a 'common sense' check will not be possible without the units.
f(t) = 250t - 25t²
1. maximum height occurs when f'(t) = 0
f'(t) = 250 - 25*2t = 250 - 50t = 0
t= 250/50 = 5
f(5)
= 250*5 - 25*5²
= 1250 - 625
=625
2. Velocity
Velocity is f'(t).
When it hits the ground, solve for t when f(t)=0.
f(t) = 250t - 25t² = 0
t(250-25t) = 0
t=0 or t=10
f'(t) = 250 - 25*2t
f'(10) = 250 - 25*2*10 = 250 - 500 = -250
(equals the initial upwards velocity of 250 in magnitude).
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(1) To find the maximum height, we need to find the vertex of the parabolic function f(t) = 250t - 25t^2.
The vertex of a parabola given by the equation f(t) = at^2 + bt + c can be found using the formula t = -b / (2a).
In this case, a = -25 and b = 250.
t = -b / (2a) = -250 / (2(-25)) = -250 / (-50) = 5.
So, the maximum height occurs at t = 5 seconds.
To find the maximum height, substitute t = 5 into the equation f(t) = 250t - 25t^2:
f(5) = 250(5) - 25(5)^2 = 1250 - 25(25) = 1250 - 625 = 625.
Therefore, the maximum height is 625 units.
(2) To find the velocity when the ball hits the ground, we need to find the time it takes for the ball to reach the ground. At the ground, the height is equal to zero.
So, we need to solve the equation f(t) = 250t - 25t^2 = 0.
Factoring out a common factor:
t(250 - 25t) = 0.
Setting each factor equal to zero:
t = 0 (which represents the starting point) or 250 - 25t = 0.
Solving 250 - 25t = 0 for t:
250 - 25t = 0
25t = 250
t = 250 / 25
t = 10.
So, the ball hits the ground after 10 seconds.
Now, to find the velocity when the ball hits the ground, we need to find the derivative of the height function f(t) with respect to time t:
f'(t) = 250 - 50t.
Substitute t = 10 into f'(t):
f'(10) = 250 - 50(10) = 250 - 500 = -250.
Therefore, the velocity when the ball hits the ground is -250 units per second.
To find the maximum height of the ball, we need to determine when the vertical velocity of the ball becomes zero. This occurs at the highest point of the ball's trajectory before it starts to fall back down.
(1) To find the maximum height, we can use the formula for vertical velocity, which is the derivative of the height function with respect to time (t).
f(t) = 250t - 25t^2
The derivative of f(t), denoted as f'(t), gives us the velocity function of the ball:
f'(t) = 250 - 50t
To find the maximum height, we set the velocity function equal to zero and solve for t:
250 - 50t = 0
50t = 250
t = 5
So, the ball reaches its maximum height at t = 5 seconds.
Now, we substitute t = 5 into the height function to find the maximum height (f(5)):
f(5) = 250(5) - 25(5^2)
f(5) = 1250 - 25(25)
f(5) = 1250 - 625
f(5) = 625
Therefore, the maximum height of the ball is 625 units.
(2) To find the velocity of the ball when it hits the ground, we need to determine the time it takes for the ball to hit the ground. This is the time at which the height function equals zero.
So we set the height function equal to zero and solve for t:
250t - 25t^2 = 0
25t(10 - t) = 0
We have two solutions: t = 0 and t = 10. However, since the ball is thrown up, we ignore the t = 0 solution.
Therefore, the ball hits the ground at t = 10 seconds.
To find the velocity at t = 10, we once again use the velocity function:
f'(t) = 250 - 50t
Substituting t = 10 into the velocity function:
f'(10) = 250 - 50(10)
f'(10) = 250 - 500
f'(10) = -250
Hence, the velocity of the ball when it hits the ground is -250 units per second (considered negative because the ball is moving downwards).