A sled with rider having a combined mass of 120 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 (in m)?

Question

Answer 1

The rider begins with

K E = \frac{1}{2} m v^{2} = 0.5 (120 k g) {(22.5 \frac{m}{s})}^{2} = 30375 J

and

P E = m g h = 120 \cdot 9.8 \cdot 0 = 0

so,

E_{t} = K E = 30375 J

where
KE=kinetic energy
PE=potential energy
E=total energy=KE+PE
At the top of the hill some of the KE is turned into PE=mgh
PE becomes

P E = 120 k g \cdot 9.8 \frac{m}{s^{2}} \cdot 11.0 m = 12936 J

and since total energy is conserved KE becomes

K E = E_{t} - P E = 30375 J - 12936 J = 17439 J

K E = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{\frac{2 \cdot K E}{m}} = \sqrt{\frac{2 \cdot 17439}{120}} = 17.05 \frac{m}{s}

Now since the hill is 11.0 m high the period of time of thevertical fall is given by

h = v_{0} t + \frac{1}{2} a t^{2}

h=11.0m

v_{0} = 0

(this is the initial vertical velocity)
a=9.8m/s
solve for t using the quadratic formula

t = \frac{- v_{a} \pm \sqrt{v_{0}^{2} - 4 (\frac{1}{2} a) (- h)}}{2 (\frac{1}{2} a)}

t = \pm \frac{\sqrt{2 a h}}{a}

choose the positive answer, as the negative answer for time is meaningless

t = \frac{\sqrt{2 (9.8) (11)}}{9.8} = 1.498 s \approx 1.5 \sec

so we know it takes 1.5 sec to hit the ground and we know the horizontal velocity is v=17.05m/s, distance travelled before impact:

d = v t = 17.05 \frac{m}{s} \cdot 1.5 s = 25.5 m

A sled with rider having a combined mass of 120 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 (in m)?

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