Question

A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 10.3 m/s. The student throws a ball along a path that she judges to make an initial angle of 50.7° 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 the ball rise? (Neglect air resistance.)

Answer 1

Answer:

8.07 m

Explanation:

The ball have two velocities , one vertically upwards v and second horizontally of 10.3 m /s. The resultant of the two makes 50.7 degree with the horizontal.

So v / 10.3 = tan 50.7

v = 10.3 tan 50.7

v = 12.58 m/s

Now v² = u² - 2gh

h = v² / 2g

= $\frac{12.58\times12.58}{2\times9.8}$

= 8.07 m