Question

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

Answer 1

Let's denote the horizontal projection as v_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 = h₀ + v_0yt - gt² / 2.
A. For our case:
X: v_x = v_0x= 42;
Y: v_y = v_0y - 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:
v_y = -10t = - 10 * 3 √ 2 = -30 √ 2 m/sec;
Using Pythagoras' theorem,
V = √ v_y² + v_x² = √ 42² + 30²*2 = 59.7 m/sec;
C. Analogically to the previous:
h = h₀ + v_0yt - gt² / 2.60 = 90 - 5t²;
5t² = 30;
t = √ 6;
v_y = -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.