Answer:
The initial velocity is 16 m/s
Explanation:
The position of the ball at a time t is given by the following vector:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
r = position vector at time t
x0 = initial horizontal position
v0x = initial horizontal velocity
t = time
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
When the ball reaches the ground its position vector will be "r final" (see attached figure). The x-component of this vector is 70 m and the y-component is -90 m. Notice that the origin of the frame of reference is located at the throwing point (y0 and x0 = 0). Then, we can use the equations for the x and y-component of "r" to obtain the time of flight and the initial velocity:
Using the equation for the y-component of the vector position:
y = y0 + v0y · t + 1/2 · g · t² (y0 and v0y = 0)
-90 m = -1/2 · 9.8 m/s² · t²
-2 · -90 m/ 9.8 m/s² = t²
t = 4.3 s
Using the equation for the x-component of the vector position:
x = x0 + v0x · t
70 m = v0x · 4.3 s
v0x = 16 m/s
The initial velocity is 16 m/s
Answer:
The applied torque is 3.84 N-m.
Explanation:
Given that,
Moment of inertia of the wheel is
Initial speed of the wheel is 0 (at rest)
Final angular speed is 25 rad/s
Time, t = 13 s
The relation between moment of inertia and torque is given by :
So, the applied torque is 3.84 N-m.