A sled and rider have a combined mass of 125kg. The sled approaches a perfectly

Question

A sled and rider have a combined mass of 125kg. The sled approaches a perfectly smooth icy hill with a speed of 22.5m/s. the top of the hill is horizonyal and it's height is 11m.

How far does the sled land from the foot of the first hill?

What is the speed of the sled just before landing?

The shape of the hill is changed as shown with =15 degrees. The height of the hill is still 11m. What is now the speed of the sled just before landing? Explain

Please show all work necessary to complete the problem. Thanks in advance!

Explanation / Answer

1/2 m*u1^2 = mgh + 1/2 m*u^2

=> 1/2 * 22.5^2 = 9.8*11 + 1/2 * u^2

=> u = 17.04 m

case 1:

u in hori zontal direction

R = u*(2h/g)

1/2 m*u1^2 = mgh + 1/2 m*u^2

=> 1/2 * 22.5^2 = 9.8*11 + 1/2 * u^2

=> u = 17.04 m in hori zontal direction

=> R = 25.54 m

case 2:

u at angle 15 degrees

R = r1+r2

r1 = u^2* sin 2 / g = 14.81m

r2 = ucos * (2h/g)

= 24.66 m

=> R = 39.47 m

case 1: speed of the sledge before landing = 22.5 m/s

since , no loss of energy occurs over the entire process

same for case 2

since speed is asked, no direction is needed

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