Abox is sliding along a frictionless surface and gets to a ramp. disregarding friction, how fast should the box be going on the ground in order to slide up the ramp to a height of 2.5 meters, where it stops? (use g = 9.8 m/s2.)

7.0 m/s

From the information;

we have the final velocity, distance, and acceleration,

We therefore, need to find initial velocity;

Using the equation;

vf^2 = vi^2 + 2ad

Making, vi² the subject;

vi² = vf² - 2ad

but; vf is zero because it comes to a stop or rest

Thus; vi² = - 2ad

vI = √(2ad) 

vi= the square root of 2ad

 = √(2×9.8 × 2.5)

vi= 7 m/s

Therefore; initial velocity = 7 m/s

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initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft