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h(t) = - 16 t 2 + 20 t + 4
h(t) = - 16 t 2 + 20 t + 4
h ( ) = 2 + +h () = ANSWER: The height of the ball at given initial snapshot of time t = 0.75 second
initial height of ball at h(t = 0) is 4 feet. How do you know it is 4 feet? You refer to the given information of the height equation; h(t) = - 16 t 2 + 20 t + 4 . By substituting in equation t = 0, you will get height , h(t=0) = 4 feet.
Imagine you measure the height from a person's hand holding the ball to the ground before he throws it upward.
and final time,
t final =
h ( ) = 2 + +
h ( ) = + +
h ( ) = final height of the ball at final given time, t = 1.25 second
At time = 0.75 seconds, the object is still at the air, the height from the ground is 10 feet.
At time = 1.25 seconds, the object is falling down, the height from the ground is 4 feet
| Time (in Seconds) | Height(in feet) | Δ Height (in feet) / | Δ Time (in seconds) | Average Rate of Change ΔH /ΔT |
| 0.75 initial | 10 initial | - | - | ANSWER (- 6 / 0.5 = -12) |
| 1.25 final | 4 final | ΔH = Hfinal - Hinitial | ΔT = Tfinal - Tinitial | minus - sign, means falling down for direction |
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