Question
A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 18.0 m/s. The student throws a ball along a path that she judges to make an initial angle of 68.0° with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise straight up. How high does the ball rise?
Answers
evidently the horizontal component of the ball's velocity is 18.0m/s toward the rear of the train.
18.0 = v*cos68.0°
v = 48m/s
The vertical component is thus 48*sin68.0° = 44.5 m/s
the height above the train is thus
h = 44.5t - 4.9t^2
= t(44.5 - 4.9t)
h is max at t=4.54
h(4.54) = 101m
18.0 = v*cos68.0°
v = 48m/s
The vertical component is thus 48*sin68.0° = 44.5 m/s
the height above the train is thus
h = 44.5t - 4.9t^2
= t(44.5 - 4.9t)
h is max at t=4.54
h(4.54) = 101m
steve where did you get 101m from