Answer:
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
Answer: T = 2.04s,
Horizontal component of velocity = 17.32m/s
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
ok confusion but we could figure it out right
It is si-32
