Projectile motion

Question

a skateboarder rolls off a horizontal ledge that is 1.32 m high, and lands 1.88 m from the base of the ledge. What was his initial velocity?

Answer 1

Hi Farrah!

 

Let's see if a suggestion, coupled with Arturo's and my answers to your previous post, may help give you a push here.

 

In this case, you should use the y direction to calculate the time the skateboarder is in the air, then take that time to the horizontal direction and solve for the skater's velocity.

 

It is worth noting that, in the horizontal direction for a projectile, the acceleration is zero (since the only defined acceleration for a projectile is gravity, which is totally in the vertical direction).  Therefore, the horizontal velocity must be constant (acceleration is related to change in velocity, and if the acceleration is zero, there must be no change in velocity).

 

Thus, in the horizontal direction, the velocity the projectile has at the start is the same velocity it has the whole time it is in flight.

 

Also, because of acceleration being zero, there is only one useful equation of motion in the horizontal, which is:

 

d = vt

 

the (perhaps familiar) "distance = rate times time"  So, since the skateboarder left the ledge with only horizontal velocity (since it was a horizontal ledge), you can just use the above equation, as long as you know the horizontal distance (given) and the time (which you solve for in the vertical direction).

 

See if that helps out, and, if you are still stuck, just let me know and we can talk about it some more.