SOLUTION: Please help me with this problem: An object is launched at 19.6 meters per second from a 58.8 meter tall platform. The equation for the object's height s at time t seconds after la
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Question 1034545: Please help me with this problem: An object is launched at 19.6 meters per second from a 58.8 meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = -4.9t2+ 19.6t + 58.8, where s is in meters.
a. When does the object strike the ground?
b. What was its maximum height?
c. How long will it take to reach its maximum height?
Found 2 solutions by Alan3354, addingup:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
An object is launched at 19.6 meters per second from a 58.8 meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = -4.9t2+ 19.6t + 58.8, where s is in meters.
a. When does the object strike the ground?
s(t) = -4.9t2+ 19.6t + 58.8
-4.9t2+ 19.6t + 58.8 = 0
Solve for t, ignore the negative solution.
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b. What was its maximum height?
It's the vertex of the parabola.
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c. How long will it take to reach its maximum height?
t of the vertex.
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
You want to know when the object hits the ground so I'm looking for the time when the height s = 0
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The formula is:
s(t) = –gt^2+v_0t+h_0
The initial launch height was 58.8 meters, and the initial velocity (launch speed) was 19.6 m/s:
0 = –4.9t2+19.6t+58.8
0 = t2–4t–12
0 = (t–6)(t+2) split into 2 equations
t-6 = 0 or t+2 = 0
t = 6 or t = -2
We know the answer we are looking for is a positive number. Discard the -2:
The object strikes the ground in 6 seconds
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