# Answer on Mechanics | Relativity Question for Dacian

Question #743

A ball is thrown at the edge of a cliff 90.0m high at 42m/s. Calculate A.) the time required to hit the ground below. B.) the velocity. C.) the velocity when the ball is 60.0m above the ground

Expert's answer

Let's denote the horizontal projection as v

X: v

h = h

X: v

Y: v

0 = 90 + 0 - 10 t

90 = 5t

t= √ 18 = 3 √ 2 sec = 4.24 sec;

v

Using Pythagoras' theorem,

V = √ v

h = h

5t

t = √ 6;

v

V = √ 42

_{x}and vertical projection as v_{y}. The equations of the motion would be:X: v

_{x}= v_{0x};Y: v_{y}= v_{0y}- gt;h = h

_{0}+ v_{0y}t - gt^{2}/ 2.**A.**For our case:X: v

_{x}= v_{0x}= 42;Y: v

_{y}= v_{0y}- gt = 0 - 10t = -10t;0 = 90 + 0 - 10 t

^{2}/2 = 90 - 5t^{2}.90 = 5t

^{2};t= √ 18 = 3 √ 2 sec = 4.24 sec;

**B.**Let's find Y-projection instantly before hitting the ground:v

_{y}= -10t = - 10 * 3 √ 2 = -30 √ 2 m/sec;Using Pythagoras' theorem,

V = √ v

_{y}^{2}+ v_{x}^{2}= √ 42^{2}+ 30^{2}*2 = 59.7 m/sec;**C.**Analogically to the previous:h = h

_{0}+ v_{0y}t - gt^{2}/ 2.60 = 90 - 5t^{2};5t

^{2}= 30;t = √ 6;

v

_{y}= -10t = - 10 √ 6 ;V = √ 42

^{2}+ 600 = 48.62 m/sec;

Answers: A. t(h=0) = 4.24 sec. B. V(h=0) = 59.7 m/sec. C. V(h=60) = 48.62 m/sec.Answers: A. t(h=0) = 4.24 sec. B. V(h=0) = 59.7 m/sec. C. V(h=60) = 48.62 m/sec.

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