The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
Answer:
The maximum height the pebble reaches is approximately;
A. 6.4 m
Explanation:
The question is with regards to projectile motion of an object
The given parameters are;
The initial velocity of the pebble, u = 19 m/s
The angle the projectile path of the pebble makes with the horizontal, θ = 36°
The maximum height of a projectile,
, is given by the following equation;

Therefore, substituting the known values for the pebble, we have;

Therefore, the maximum height of the pebble projectile,
≈ 6.4 m.