- #1
pooface
- 208
- 0
TEST question!
Golf ball is struck at 12 m above the plain of a tee giving it a velocity of 40m/s at an angle of 30degrees.
a) find the time of flight
b) find max height above plane
c) horizontal distance to point of landing.
d) final velocity and angle.
Vox = 40cos30
Voy = 40sin30
a) -12 = 40sin30t + 0.5(-g)t^2
t=4.6084 seconds
b) disregarding the 12m elevation (which i will add later on).
0=40sin30t + 0.5(-g)t^2
t=4.0775s
t/2 = time at peak
y= - 0.5(-g)t^2 +12
= 32.387m
c)x =voxt
= 34.34(4.6084)
=158.25m
d)
-12= Vfy(4.6084) -0.5(-g)(4.6084)^2
Vfy = 25.21m/s
angle = arctan(vfy/vox) = -36.28deg
Homework Statement
Golf ball is struck at 12 m above the plain of a tee giving it a velocity of 40m/s at an angle of 30degrees.
a) find the time of flight
b) find max height above plane
c) horizontal distance to point of landing.
d) final velocity and angle.
Homework Equations
The Attempt at a Solution
Vox = 40cos30
Voy = 40sin30
a) -12 = 40sin30t + 0.5(-g)t^2
t=4.6084 seconds
b) disregarding the 12m elevation (which i will add later on).
0=40sin30t + 0.5(-g)t^2
t=4.0775s
t/2 = time at peak
y= - 0.5(-g)t^2 +12
= 32.387m
c)x =voxt
= 34.34(4.6084)
=158.25m
d)
-12= Vfy(4.6084) -0.5(-g)(4.6084)^2
Vfy = 25.21m/s
angle = arctan(vfy/vox) = -36.28deg
Last edited: