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PROJECTILE
A golf ball was hit and projected at an angle of 60 degrees horizontal. If the initial velocity of the ball is 50 m/s; Calculate:
a. the total time the golf ball was in the air?
b. the horizontal distance?
c. its maximun height?
I have my solution. I will choose the best answer similar to my answer.
the starting vertical speed of the ball = 50 Sin 60 = 25√3 m/s
horizontal speed is constant through out the motion is constant because there is no unbalanced force exerting on the ball horizontally
and it’s = 50 Cos 60 = 25 m/s
starting speed = end speed because of conservation of Mechanical Energy but vertical speed is constant so the vertical starting velocity = -(vertical end velocity)
mutiplied by – because velocity is vector and in this case it’s in the opposite direction to the start velocity
to the ball vertically till it reach ground downwards (+)
v = u + at
25√3 = – 25√3 + gt
t = 50√3 /g
t = 50 * 1.732/9.8
t = 8.84 s
to the object horizontally
S = VT
S = 50√3 * 25 / g
S = 1250 * 1.732 / 9.8
S = 220.92 m
to the object vertically till the vertical speed=0 , upwards (+)
v² = u² +2as
0 = (25√3)² – 2gh
h = (625 * 3) / (2 * 9.8)
h = 1875 */ 19.6
h = 95.66 m
