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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.

User Antonio
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1 Answer

3 votes

Answer:

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

User MySun
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