Asked by Samuel
A diver on a platform 40 feet in height jumps upward with an initial velocity of 5ft/s. His height in h feet after t seconds is given by the function h=-16t^2+5t+50.
a. What is his maximum height? I got 50.39
b. How long will it take him to reach the surface of the water?
h=-16t^2+5t+50
50.39 = -16t^2+5t+50
-16t^2+5t+50 = 50.39
I am going to try completing the square.
(-16t^2/16)+(5/16)t = 0.39/16
t^2 + (5/16)t = .02
I'm stuck.
a. What is his maximum height? I got 50.39
b. How long will it take him to reach the surface of the water?
h=-16t^2+5t+50
50.39 = -16t^2+5t+50
-16t^2+5t+50 = 50.39
I am going to try completing the square.
(-16t^2/16)+(5/16)t = 0.39/16
t^2 + (5/16)t = .02
I'm stuck.
Answers
Answered by
Jai
Is the platform really 40 feet in height, not 50 feet...? Anyway,
a. Yes, that value is right.
b.
Time it takes from platform to maximum height (I used derivatives):
dh/dt = -32t + 5
0 = -32t + 5
t = 0.15625 s
Time it takes from maximum height to surface of water:
h = vo*t - (1/2)*g*t^2
But vo = 0 because it's freefall, and g = 32.2 ft/s^2:
50.39 = 0 - (0.5)*(-32.2)*(t^2)
50.39 = 16.1(t^2)
t^2 = 3.1298
t = 1.769 s
Total time: 1.769 + 0.15625 = 1.925 s
hope this helps~ `u`
a. Yes, that value is right.
b.
Time it takes from platform to maximum height (I used derivatives):
dh/dt = -32t + 5
0 = -32t + 5
t = 0.15625 s
Time it takes from maximum height to surface of water:
h = vo*t - (1/2)*g*t^2
But vo = 0 because it's freefall, and g = 32.2 ft/s^2:
50.39 = 0 - (0.5)*(-32.2)*(t^2)
50.39 = 16.1(t^2)
t^2 = 3.1298
t = 1.769 s
Total time: 1.769 + 0.15625 = 1.925 s
hope this helps~ `u`
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