quick math question

Question

1. A golf ball is hit and lands 53 m away. The path of the ball took it just over a 9 m tree. Determine a function that models the path of the golf ball.

Answer 1

let 9 = max height

h(t) = -4.9^2 + vt -4.9 = a/2 where a= -9.8 m/sec^2, v = initial upward velocity

h'(t) = -9.8t + v = 0

t= v/9.8 = time to reach max height 9

9= -4.9(v/9.8)^2 + v(v/9.8)= -v^2/19.6 + v^2/9.8= v^2/19.6

v^2/19.6 = 9

v^2 =(9)(19.6

v = 3sqr19.6 = about 13.28 m/sec

h(t)=-4.9t^2 + (3sqr19.6)t = height of the ball at time t

at t=3sqr19.6/9.8 = about 1.355 seconds, the ball reaches maximum height = 9 meters

at t=2(3sqr19.6)/9.8 = about 2.71 seconds, the ball lands 53 meters away

r(t) = 53/2.71 = 19.56 meters per second = horizontal velocity

d(t) =rt = 19.56t = horizontal distance of the ball at time t

position of the ball at time t = (x,y) = (d,h) = (19.6t, -4.9t^2+13.3t) where

0<t<2.71

0<d<53,

and 0<h<9

(assuming the ball began at h=0, ground level and landed at ground level)

(oddly if you used deceleration due to gravity of -16t^2 +vt, v would equal exactly 24 ft/sec with time at maximum height of 3/4 sec. if 9 were in feet, not meters. It's as if someone mixed up feet and meters in this problem)