Question

A ball is projected horizontally above level ground from the top of a vertical cliff. the ball strikes the level ground 1 km away from the base of the cliff 4 seconds after it is fired. the height of the cliff is:

a) 80
b) 160
c) 250

Answer 1

(c) 250

Step-by-step explanation:

We can start by analyzing the horizontal and vertical motion of the ball separately. Since the ball is projected horizontally, its initial vertical velocity is 0.

Horizontal motion:

Initial horizontal velocity (Vx) = distance/time = 1 km / 4 s = 250 m/s

Final horizontal velocity (Vx) = same as initial velocity, since there is no horizontal acceleration

Vertical motion:

Initial vertical velocity (Vy) = 0

Vertical acceleration (a) = -9.8 m/s^2 (assuming downward direction as negative)

Time of flight (t) = 4 s

Final vertical velocity (Vy) = Vy + a*t = -9.8 m/s^2 * 4 s = -39.2 m/s

Vertical displacement (h) = Vyt + 0.5at^2 = 0 + 0.5(-9.8 m/s^2)*(4 s)^2 = -78.4 m (negative because the ball is falling below the initial height)

Now, we can use the Pythagorean theorem to find the height of the cliff:

h^2 + d^2 = (78.4 m)^2

h^2 + (1000 m)^2 = (78.4 m)^2

h^2 = (78.4 m)^2 - (1000 m)^2

h^2 = 61344 m^2

h = sqrt(61344 m^2) = 248 m

Therefore, the height of the cliff is approximately 248 m. The closest option is (c) 250.