Question

A skateboarder jumps horizontally off the top of a staircase at a speed of 14.5 – and lands at bottom of the

Answer 1

Question:

A skateboarder jumps horizontally off the top of a staircase at a speed of 14.5 and lands at bottom of the stairs. The staircase has a horizontal length of 8.00 m. We can ignore air resistance. What is the skater's vertical displacement during the jump?

Answer:

y = 1.48 m

Explanation:

Projectile motion is a two dimensional motion experienced by an object or particle that is subjected near the Earth's surface and moves along a curved path under the influence of gravity only. The path followed by projectile motion is called projectile path.

As the skateboarder followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

Also Skateboard jumps horizontally, So initial velocity has only horizontal component.

Horizontal component of initial velocity = $v_{i_{x}}$ = 14.5 m/s

Horizontal displacement = x = 8.00 m

Vertical displacement = y = ?

Using the following formula

x = $v_{i_{x}}$t

8 = (14.5)(t)

t = 0.55 s

As skateboarder jumps horizontally, So there is no vertical component of velocity.

According to 2nd equation of motion

y = $v_{i_{y}}t + \frac{1}{2}gt^{2}$

As $v_{i_{y}} = 0$

So

y = 0.5gt²

y = 0.5*9.8*(0.55)²

y = 1.48 m

Answer 2

Answer: -1.49

Explanation:

/\x = vxt

t = /\x / vx

t = 8m / 14.5 m/s

= 0.552s

/\y = v0y + 1/2 ay t^2

= (0)t + 1/2 (-9.8 m/s^2) (0.552s)^2

= -1.49m (answer)

:))) This is the actual answer