Answer:
11.09 m/s
Explanation:
Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.
The parameters given are:
Initial velocity U = ?
Final velocity V = 9.6 m/s
Acceleration due to gravity g = 9.8m/s^2
Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0
Using third equation of motion
V^2 = U^2 - 2gH
0 = U^2 - 2 × 9.8H
U^2 = 19.6H ........ (1)
Using the formula again for one fourth of its maximum height
9.6^2 = U^2 - 2 × 9.8 × H/4
92.16 = U^2 - 19.6/4H
92.16 = U^2 - 4.9H
U^2 = 92.16 + 4.9H ...... (2)
Substitute U^2 in equation (1) into equation (2)
19.6H = 92.16 + 4.9H
Collect the like terms
19.6H - 4.9H = 92.16
14.7H = 92.16
H = 92.16/14.7
H = 6.269
Substitute H into equation 2
U^2 = 92.16 + 4.9( 6.269)
U^2 = 92.16 + 30.72
U^2 = 122.88
U = 11.09 m/s
Therefore, the initial velocity of the object is 11.09 m/s
Answer: the pitch would be higher than it was at a lower speed because the air in front of the jet is compressed
Answer:
= 351.84 J
Explanation:
Using the conservation of energy K:
so:
where m is the mass, v the initial velocity, is the kinetic energy of the mass as it clears the fence, g the gravity and h the altitude.
Then, replacing values, we get:
solving for :
= 351.84 J
Answer:
B) changing position
Explanation:
When a ball bounces to the ground it hits the ground with some energy. The amount of energy with which it hits the ground is kinetic energy. When it comes in the contact with the ground kinetic energy gets converted into potential energy. This potential energy again gets converted into kinetic energy and balls moves again from the ground and bounces multiple times. So, due to multiple bounce the position of the ball changes.
Thus, When bouncing a ball, the bouncing motion results in the ball changing position.
