Answer to Question #743 in Mechanics | Relativity 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
1
Expert's answer
2010-10-10T17:57:44-0400
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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