PreCalculus
posted by Lucina .
1. A projectile is fired straight up from a height of 6 feet. Its height (h) in feet after t seconds is given by h = 6 + 192t 16t^2.
Answer:
h = 16t^2 + 192t + 6
h = 16( t^2 12t +36) + 6 +576
h = 16 (t6)^2 +582
The maximum height will be 582 feet.
2. The path of a basketball on the moon can be modelled by the equation h(d) 9d^2+90d+2 where h(d) is the height of the basketball in meters and d is the horizontal distance of the ball from the player. Dtermine
a) height of ball before it is show
h(0) = 2
The height of the ball is 2 m
b) The maximum height
h(d) = 9 ( d^2  10d + 25) + 2 + 225
h(d) = 9 ( d5)^2 + 227
The maximum height is 227 m
c) How long will it take the ball to reach the maximum height?
Since the vertex is (5,227), it will take 5 seconds
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