A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo
1 answer:
Answer:
Explanation:
Data provided in the question:
Height above the ground, H= 5.0m
Range of the ball, R= 20 m
Initial horizontal velocity =
Initial vertical velocity= (Since ball was thrown horizontally only)
Acceleration acting horizontally, = 0 m/s² [ Since no acceleration acts horizontally) ]
Vertical Acceleration, = 9.8 m/s² (Since only gravity acts on it)
Let 't' be the time taken to reach ground
Therefore, using equations of motion, we have
Then using Equations of motion for horizontal motion,
You might be interested in
The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.