Question

The top of the ramp is h1 = 1.17 m above the ground; the bottom of the ramp is h2 = 0.298 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance d away. Calculate the distance d.

Answer 1

d = 1.02 m

Step-by-step explanation:

By energy conservation we can find the speed by which ball will leave the ramp

`U_i + KE_i = U_f + KE_f`

here we know that

`mgh_1 + 0 = mgh_2 + (1)/(2)mv^2`

here we have

`h_1 = 1.17 m`

`h_2 = 0.298 m`

so we have

`(9.8)(1.17) = (9.8)(0.298) + (1)/(2) v^2`

`v = 4.13 m/s`

now the time taken by the block to reach the ground is given by

`h_2 = (1)/(2)gt^2`

`t = \sqrt{(2h_2)/(g)}`

`t = \sqrt{(2(0.298))/(9.8)}`

`t = 0.25 s`

now the distance covered by it is given as

`d = 0.25 * 4.13`

`d = 1.02 m`