Baseball is thrown up in the air from a height of 3 feet with an initial velocity of 23 feet per second. what is the maximum height of the baseball and when does it reach this height? [use the equation h = -16t2 + v0t + h0.] a) the maximum height is 7.2 feet and it reaches this height at 0.85 seconds. b) the maximum height is 11.3 feet and it reaches this height at 0.72 seconds. c) the maximum height is 2.35 feet and it reaches this height at 29.98 seconds. d) the maximum height is 135.25 feet and it reaches this height at 11.50 seconds.

B) The maximum height is 11.3 ft and it reaches this height at 0.72 seconds.

Explanation:
Part (a) : getting the time taken to reach maximum height:
The given equation is:
h = -16t² + V0*t + h0

We are given that:
Initial velocity (V0) =23 ft/sec
Initial height (h0) = 3 ft

The equation now becomes:
h = -16t² + 23t + 3

For maximum height:
dh / dt = zero
This means that:
-16(2) t + 23 = 0
-32t + 23 = 0
32t = 23
t = 0.71875 sec which is approximately 0.72 seconds

Part (b): getting the maximum height:
The equation given to get the height is:
h = -16t² + V0*t + h0

We are given that:
Initial velocity (V0) =23 ft/sec
Initial height (h0) = 3 ft

We have calculated that:
time taken to reach maximum height = 0.72 sec

Substitute with the givens in the equation to get the maximum height as follows:
h = -16t² + V0*t + h0
h = -16(0.72)² + 23(0.72) + 3
h = 11.2656 ft which is approximately 11.3 ft

Hope this helps :)

answer: 2530

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