Question

A ball was thrown from a projectile building of 30m which moves at a constant velocity of 20m/s and has an angle of 30degrees to the horiontal .Calculate the time of flight to the ground. And th e horizontal components

Answer 1

Answer: T = 2.04s,

Horizontal component of velocity = 17.32m/s

Step-by-step explanation:

Height (h) = 30m

Initial velocity (u) = 20m/s

Theta = 30°

g = acceleration due to gravity = 9.8m/s^2

Time of flight(T), time taken by a projectile to return back to the ground

T = (2*u*sin(theta)) ÷ g

T = (2*20*sin(30)) ÷9.8

T = (40 * 0.5) ÷ 9.8

T = 20 ÷ 9.8

T = 2.04s

Horizontal component of velocity describes the Influence of velocity in displacing the projectile horizontally.

Let Vx = horizontal component of velocity

Using SOHCAHTOA,

CAH = cos = adjacent/hypotenus

Cos 30° = Vx / 20

Vx = 0.866 * 20

Vx = 17.32m/s

Answer 2

Time of flight = 4.08seconds

Horizontal component of initial velocity is 17.32m/s

Explanation: complete question( and the horizontal component of the initial velocity.)

The equation for time of flight of a projectile is given as T= 2u/g

T=( 2×20)/9.8

T= 40/9.8= 4.08seconds

Horizontal component of initial velocity Vix= Vi Costheta

Vix= 20× cos 30°

Vix= 20×0.8660

Vix= 17.32m/s