# 2d motion kinematics (phys) probs

• Feb 7th 2007, 04:22 AM
thedoge
2d motion kinematics (phys) probs
These aren't especially hard, but I have no time to examine them since being assigned as I've had stuff to deal with yesterday. Now time is running short. Any help would be appreciated.

A skier leaves the ramp of a ski jump with a velocity of v = 18.0 m/s, = 15.0° above the horizontal. The slope is inclined at 50.0°, and air resistance is negligible. (Assume up and right are positive, and down and left are negative.)

Find total distance from edge of ramp to landing.
Find velocity components just before landing.

A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 8.50 m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 50.0° 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 vertically. How high does she see the ball rise?
• Feb 7th 2007, 10:33 AM
ticbol
Quote:

Originally Posted by thedoge
These aren't especially hard, but I have no time to examine them since being assigned as I've had stuff to deal with yesterday. Now time is running short. Any help would be appreciated.

A skier leaves the ramp of a ski jump with a velocity of v = 18.0 m/s, = 15.0° above the horizontal. The slope is inclined at 50.0°, and air resistance is negligible. (Assume up and right are positive, and down and left are negative.)

Find total distance from edge of ramp to landing.
Find velocity components just before landing.

A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 8.50 m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 50.0° 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 vertically. How high does she see the ball rise?

Cannot quite understand the "= 15 degrees above horizontal" in the first problem. You mean that's the angle of the jump?

For the 2nd problem, since there is no more data to play with than the givens, I assume the velocity of the ball is due to the motion of the train only. The ball leaves the thrower's hand with a velocity whose horizontal component is the velocity of the train.

Since the we are asked to find how high the ball goes up, then we are interested on the vertical component of the velocity of the ball as the ball leaves the student's hand.

Let v = that vertical component, in m/sec.
And H = maximum height attained by the ball above the tha raised hand of the student, in meters.
t = time for the ball to reach H, in seconds.

tan(50deg) = v/8.5
v = 8.5tan(50deg) = 10.13 m/sec.

H = v*t -(1/2)g*t^2 ------------(i)

To find t, we know that at H, the ball stops moving up, so,
v -gt = 0
t = v/g = 10.13/9.8 = 1.03367 sec.
Substitute that into (i),
H = 10.13*1.03367 -(1/2)(9.8)(1.03367^2) = 5.23556 m.

Therefore, the lady Prof sees the ball rise a maximum height of about 5.24 meters above the palm of the student. --------answer.