A golf ball is hit an angle of 35 degrees above the horizontal and it lands in a hole 120 m away 4.2 s later. Air resistance is negligible.?

Question

a) What was the initial speed of the golf ball? Include a labelled diagram.
b) Determine the maximum height the ball reached. (*If you were not able to answer part (a), then assume the ball's initial speed is 25 m/s) (3 marks)

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Answer 1 · 2018-03-22T06:23:27

a) 35m/s
b) 22m

a) In order to determine the initial velocity of the golf ball I found the x and y components.

Since we know that it travelled 120m in 4.2s we can use this to calculate initial x velocity

initial Vx = $\frac{120 m}{4.2 s}$ $(120m)/(4.2s)$ = 28.571m/s.

To find the initial y velocity we can use the formula

$d = V i (t) + \frac{1}{2} a t^{2}$ $d=Vi(t)+1/2at^2$

We know that the y displacement = 0 after 4.2s so we can plug in 0 for d and 4.2s for t.

$0 = V i (4.2) + \frac{1}{2} (- 9.8) ({4.2}^{2})$ $0 = Vi(4.2)+1/2(-9.8)(4.2^2)$

Initial Vy = 20.58

Since we now have the x and y components we can use $a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$ to find the initial velocity.

${20.58}^{2} + {28.571}^{2} = V i$ $20.58^2+28.571^2 = Vi$
$V i = 35.211 =$ $Vi = 35.211 =$ 35m/s

b) To find the maximum height reached by the golf ball we can use the formula

$V f^{2} = V i^{2} + 2 a d$ $Vf^2 = Vi^2 + 2ad$

Since we know that the ball won't have any y velocity at its' maximum height we can substitute 0 for Vf and 20.58 for Vi.

$0^{2} = {20.58}^{2} + 2 (- 9.8) d$ $0^2 = 20.58^2+2(-9.8)d$
$d = 21.609 = 22 m$ $d = 21.609 = 22m$