Question

A skateboarder shoots off a ramp with a velocity of 5.5 m/s, directed at an angle of 53° above the horizontal. The end of the ramp is 1.3 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? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp? slader

Answer 1

a) 2.282 m

b) 1.481 m

Step-by-step explanation:

Given:

Initial velocity, V = 5.5 m/s

thus,

component of velocity in y-direction,
`V_y` = V sin53°

= 5.5 × sin53°

= 4.39 m/s

Component of velocity in x - direction, Vₓ = V cos58°

= 5.5 × cos53°

= 3.31 m/s

Also,

Final velocity in y direction,
`V'_y` = 0 m/s ( at the highest point )

Now,

From Newton's equation of motion

`V'_y^2 = V_y^2 + 2a_yy`

here, y is the distance travelled in y-direction

`a_y` is the acceleration in y direction i.e acceleration due to gravity 'g = -9.81'

here negative sign depicts the direction of acceleration is opposite to the direction of motion

or

y =
`(V'_y^2 - V_y2 )/(2a_y)`

on substituting the respective values, we get

y =
`(0^2 - 4.39^2 )/(2*(-9.81))`

or

y = 0.982

Hence,

the total height above the ground = y + height of ramp

= 0.982 + 1.3

= 2.282 m

b)

From Newton's equation of motion

`V'_y=V_y+at`

here t is the time

thus,

`0=4.39+(-9.81)t`

or

t = 0.447 seconds

Also,

Distance = Speed × time

thus,

Horizontal distance = Vₓ × t

or

Horizontal distance = 3.31 × 0.447

or

Horizontal distance = 1.481 m