Question

A skateboarder shoots off a ramp with a velocity of 6.9 m/s, directed at an angle of 62° above the horizo.......?

A skateboarder shoots off a ramp with a velocity of 6.9 m/s, directed at an angle of 62° above the horizontal. The end of the ramp is 1.4 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.

(a) How high above the ground is the highest point that the skateboarder reaches?

_ m

(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

_ m

Answer 1

Here, the skater follows a projectile motion, see the diagram,

so, maximum height is given as `H= (u^2 sin ^2theta) /(2g)` (using `v^2=u^2-2gH` ,here, `u=u sin theta`,and `v=0` at highest point as after that point it starts falling down due to gravity)

So, here, `u=6.9 m/s` and `theta=62`, so `H=1.85`

Here, `H` is the height up to which the skater moves above the ramp,so he will be `(1.4+1.85)m` or,`3.25 m` vertically above the ground

from the diagram, `FD` is the distance between the end point of the ramp and the highest point.

Now, `FG` = half the range of projectile = `1/2(u^2 sin 2theta)/(g)`=`1.97 m`

So, `FD` = `sqrt(FG^2+DG^2)` = `2.70 m` (from triangle,`FGD`)