Question

A golf ball is hit from 4.3 m above a golfing fairway with an initial velocity of 30.0 m/s at an angle of 35° above the horizontal.

Answer 1

$$Solution$$

$Given:\ y_D=4.3m\\ V_i=30m\ \ /s\ and\ \theta=35\degree\\ Where=V_{ix}=V_i\ cos\theta=(30m/s)\times\ Cos\ (35)\degree=24.57m/s\\ V_(iy)=V_i\ sin\theta=30m/s\times\ sin(35\degree)=17.207m/s$

i) In case we want to determine the time of flight of the ball then this can be solved as below;

Using the equation

$y=y_i+V_{iy}t+\frac{1}{2}a_yt^2\\ \therefore\ 0=4.3+(17.207)t+\frac{1}{2}(-9.81)t^2\\ (4.905)t^2-(17.207)t-4.3=0\\ Solving\ the\ quadratic\ equation\ for\ t\ gives\ t=3.742\ sec\\ ii) Also,\ the\ range\ of\ the\ golf\ ball\ can\ be\ determined\ as\ follows\\ \Delta\ d_x=V_{ix}\times\ t=(24.57m/s)(3.742sec)\\ \therefore\ \Delta\ d_x=91.94m\\$

iii) We can equally determine the velocity for the golf ball the instant before the ball impacts the ground by;

$Here,\ V_{fx}=V_{ix=24.57m/s}\\ while,\ V_{iy}+a_yt=(17.207)+(-9.81)(3.742)=-19.5m/s\\ Therefore\ V_f=\sqrt(24.57)^2+(-19.5)^2\\ =31.37m/s$