Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
On dividing equation (1) by (2), we have
Now, putting this value in equation (2), we have
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.
You need to know the speed and direction of object
Answer:
energy is equal to 1000 J
Explanation:
When the jumper is in the tent, he has a given height, this height gives him a gravitational potential energy, which forms his initial mechanical energy of 1000 J. After jumping, this energy is converted into elastic energy of the rope plus a remainder of potential energy gravitational, it does not reach the ground, but as the friction is negligible the total mechanical energy is conserved, therefore its energy is equal to 1000 J
This is a case of energy transformation, but the total value of mechanical energy does not change
Answer:
I'm just in jss2 but I read physics. this is what I think
