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<h2>The angle of the initial velocity with respect to the horizontal is 85.14°</h2>
Explanation:
Given that the ball takes 3 seconds to reach its maximum height.
Consider the vertical motion of ball till maximum height.
We have equation of motion v = u + at
Acceleration, a = -9.81 m/s²
Final velocity, v = 0 m/s
Time, t = 3 s
Substituting
v = u + at
0 = u + -9.81 x 3
u = 29.43 m/s
Initial vertical velocity is 29.43 m/s.
Now consider horizontal motion of ball.
Time of flight of ball = 2 x Time to reach maximum height = 2 x 3 = 6 s
Displacement = 15 m
We have equation of motion s = ut + 0.5 at²
Displacement, s = 15 m
Acceleration, a = 0 m/s²
Time, t = 6 s
Substituting
s = ut + 0.5 at²
15 = u x 6 + 0.5 x 0 x 6²
u = 2.5 m/s
Initial horizontal velocity is 2.5 m/s
Let r be the initial velocity and θ be the angle with horizontal
Initial vertical velocity = rsinθ = 29.43 m/s
Initial horizontal velocity = rcosθ = 2.5 m/s
Dividing
The angle of the initial velocity with respect to the horizontal is 85.14°