Question

A sled with rider having a combined mass of 110 kg travels over the perfectly smooth icy hill shown in the accompanying figure. How far does the sled land from the foot of the cliff?

Answer 1

Find the speed the sled has at the top of the hill from the law of conservation of mechanical energy. Equate the kinetic energy at the bottom of the hill to the kinetic plus potential energy at the top : 

0.5mv₀² = 0.5mv² + mgh 
v = √[v₀² - 2gh] 
= √[(22.5m/s)² - (2 x 9.80m/s² x 11.0m)] 
= 17.0m/s  

From the time independent kinematics equation, find the vertical component of the sleds final velocity (note that the vertical component of the sleds initial velocity is zero) : 

v² = v₀² + 2gΔy 
= 0 + 2(-9.80m/s²)(-11.0m) 
= -14.7m/s (select the neg root, because motion is downward)  

With this you can find the time vertically which is the same horizontally : 

v = v₀ + gt 
t = (v - v₀) / g  
= (-14.7m/s - 0) / -9.80m/s² 
= 1.50s 

Now, the horizontal distance is : 

Δx = (v₀ + v)t / 2 
= (17.0m/s + 17.0m/s)1.50s / 2 
= 25.5m 

  tell me if you need more help :)

Answer 2

The sled land is "25.531 m" far from the foot of the cliff.

According to the question,

The velocity of the top = $V_t$

then,

→ $0.5 \ Vt^2-0.5 \ Vi^2 = -mgh$

→                 $0.5 \ Vt^2 = 0.5 \ Vi^2- 9.8\times 11$  

→                 $0.5 \ Vt^2 = 0.5 \ Vi^2- 107.8$

→                              $=17.04 \ m/sec$

Now,

The projectile motion in horizontal direction,

→ $S_y = 11 = ut-0.5 \ gt^2$

                $= 0-0.5\times 9.8 \ t^2$

or,

→ $S_x = 17.04\times 1.498$

       $= 25.53 \ m$

Thus the above answer is the correct one.

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