Asked by matrix school 2
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.
(1) find the maximum height
(2) find the velocity when it hits the ground.
Answers
Answered by
MathMate
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).
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).
Answered by
junior
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