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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 vx and vertical projection as vy. The equations of the motion would be:
X: vx = v0x;Y: vy = v0y - gt;
h = h0 + v0yt - gt2 / 2.
A. For our case:
X: vx = v0x= 42;
Y: vy = v0y - gt = 0 - 10t = -10t;
0 = 90 + 0 - 10 t2 /2 = 90 - 5t2.
90 = 5t2;
t= √ 18 = 3 √ 2 sec = 4.24 sec;
B. Let's find Y-projection instantly before hitting the ground:
vy = -10t = - 10 * 3 √ 2 = -30 √ 2 m/sec;
Using Pythagoras' theorem,
V = √ vy2 + vx2 = √ 422 + 302*2 = 59.7 m/sec;
C. Analogically to the previous:
h = h0 + v0yt - gt2 / 2.60 = 90 - 5t2;
5t2 = 30;
t = √ 6;
vy = -10t = - 10 √ 6 ;
V = √ 422 + 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.

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