Nor can I.
time in air: 4.2sec
horizonantal velocity=91.4/4.2=21.8m/s
but velocityhorizontal=VcosTheta
Now, vertical
hf=hi+VsinTheta*t-4.9t^2
hf=hi=0
so t=VsinTheta/4.9 but t=4.2sec
VsinTheta=4.2*4.9 and we know
VcosTheta=21.8
dividing the equations
tanTheta=4.2*4.9/21.8
then you can solve for theta
Now for velocity. if you know theta, then
V=21.8/cosTheta
A golf ball is struck with a five iron on level ground. It lands 91.4 m away 4.20 s later. What was (a) the direction and (b) the magnitude of the initial velocity?
angle=tan-1(Ry/Rx) for direction
R=(Rx^2+Ry^2)^(1/2)
I cannot figure out the variables and if the x and y component is needed for this problem. Please help!
3 answers
For theta a got 21 and 69 as answers and they were correct. When I then plugged it back in .53/cos theta for time I got .54 and 1.4 seconds but it says these answers are wrong. Do you know where I went wrong?
Time for what? I have no idea what you are looking for. You have the time of flight already.
1 answer