Two rocks are thrown off the edge of a cliff that is 15.0 m above the ground. Th

Question

Two rocks are thrown off the edge of a cliff that is 15.0 m above the ground. The first rock is thrown upward, at a velocity of +12.0 m/s. The second is thrown downward, at a velocity of ?12.0 m/s. Ignore air resistance. Determine (a) how long it takes the first rock to hit the ground and (b) at what velocity it hits. Determine (c) how long it takes the second rock to hit the ground and (d) at what velocity it hits.

Explanation / Answer

S = 15.0 m Ua = 12.0 m/s. Ub = ?12.0 m/s. (a) how long it takes the first rock to hit the ground S = Ut - 0.5 g t^2 = 15 = 12t - .5 * 9.81 *t^2 Solve for t (b) at what velocity it hits. V = u -gt = 15 - 9.81 t (c) how long it takes the second rock to hit the ground S = U t - 0.5 g t^2 = 15 = -12t - .5 * 9.81 *t^2 Solve for t (b) at what velocity it hits. V = U - gt = -15 - 9.81 t for rock which is thrown upward ; S= +ut - gt^ 2/2 so 15 = 12 t - 4.9 t^2 => 4.9 t^2 -12 t -15=0 , t = 3.36 sec for rock which is thrown downward ; S= +ut - gt^ 2/2 so 15 = -12 t + 4.9 t^2 => 4.9 t^2 -12 t -15=0 , t = 3.36 sec same for both the rock b) v^2 = u^2 + 2 as => 12^2 + 2 *9.8 * 15 = 13.36 m/s

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