Question

Consider again the cannonball launched by a cannon from the top of a very high cliff. Suppose that the cannonball is launched horizontally with no upward angle whatsoever and with an initial speed of 20 m/s. If there were no gravity, the cannonball would continue in motion at 20 m/s in the horizontal direction. Yet in actuality, gravity causes the cannonball to accelerate downwards at a rate of 9.8 m/s/s. This means that the vertical velocity is changing by 9.8 m/s every second. If a vector diagram (showing the velocity of the cannonball at 1-second intervals of time) is used to represent how thex- and y- components of the velocity of the cannonball is changing with time, then x- and y-velocity vectors could be drawn and their magnitudes labeled. The lengths of the vector arrows are representative of the magnitudes of that quantity. Such a diagram is shown below -20m/s Vy = 20m/s Vy = 20m/s x-20m/s

Answer 1

vertical velocity just before landing = -39.2 - 9.8 = - 49 m/s

let height of cliff = H

=> v^2 = u^2 + 2aH

=> (-49)^2 = 0 + 2*9.8*H

=> H = 122.5 m

time taken to reach to ground = t

=> v = u + at

=> 49 = 0 + 9.8*t

=> t = 5 sec

NOW distance travelled horizontally = 20*t = 20*5 = 100m