Answer:
I'll assume the ? is supposed to be t^2
See the answers below.
Step-by-step explanation:
h(t) = -4.91t^2 + 24.5t + 1
a) How high is the ball after 1 second?
h(1) = -4.91*(1)^2 + 24.5*(1) + 1
h(1) = 20.59 meters
b) Find the maximum height of the ball to one decimal place.
We can find the maximum height by graphing or by taking the first derivative and setting it equal to zero (the slope is zero at the top of the curve):
First Derivative
h(t) = -4.91*(t)^2 + 24.5*(t) + 1
'h(t) = -9.82*(t) + 24.5
0 = -9.82*(t) + 24.5
9.82t = 24.5
t = 2.49 seconds is the time the ball reaches maximum height. Use that is the equation to find the height:
h(2.49) = -4.91*(2.49)^2 + 24.5*(2.49) + 1
h(2.49) = 31.6 meters
Graph
See attached graph. The maximum height occurs at t = 2.49 seconds and the height is 31.6 meters.
c) When does the ball reach its maximum height?
From the first derivative calculation above, t = 2.49 seconds to reach maximum height. The graph also shows 2.49 seconds. Better hurry.
d) When does the ball hit the ground?
The can be determined by either direct calculation of from the graph (attached).
Calculation:
h(t) = -4.91t^2 + 24.5t + 1
0 (ground) = -4.91t^2 + 24.5t + 1
4.91t^2 - 24.5t - 1 = 0 Solve with the quadratic equation. I get t = 5.04 seconds
Graph:
The graph shows 4.95 seconds. Both agree when rounded to one decimal point: 5.0 seconds.
