Question

A metal marble is launched with a speed of v0 = 30.0m/s at 5.0 ◦ relative to the horizontal, from a height of 1.0 m above ground, and towards a very tall vertical wall, which is 6.0 m away from the launch position. The projectile may first hit the wall, or the ground. Whichever it turns out to be, assume that this is an elastic, specular collision, so that the projectile bounces off. After some time, the projectile will have another collision, which again could be either with the ground or with the wall.

Find: (a) the location of the second collision; (b) the highest point (both the x- and the y-coordinates) that the projectile reaches between its launch and the second collision; (c) the velocity (the magnitude and the angle relative to the vertical—make a sketch so that it is clear which angle you are reporting) 0.50 s after launch.

Answer 1

vertical speed of ball= 30sin5^o

=2.61m/s

horizontal speed=30cos5^o

=29.9m/s

time to hit the ground just considering the vertical speed (t)

s=ut+1/2 at²

1=-2.61t+5t²

t=0.78s

Distance ball would travel horizontally in same time=29.9*0.78

=23.322m

which is way greater than 6m. So, ball hits the wall first.

time to do so = 6/29.9=0.2s

Also, if you think carefully, only the horizontal component of velocity changes its direction after collision with the wall.

So, the ball will actually continue as it would without the wall, but in opposite direction.

Our problem is simplified now.

Second collision will be with the ground at 23.22m.

At the highest point, vertical velocity=0.

0=-2.61+10t

t=0.261s

s=2.61(0.261)-5(0.261)²

=0.34m

So, height from ground=1.34m and 29.9*0.261=7.8m ahead of launch.

vertical velocity after 0.5s, v2=-2.61+10(0.5)

=2.31m/s

And this instant is after 1st collision

velocity vector is [(-29.9)i-(2.31)j]m/s