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:
An object with negative acceleration could be speeding up, and an object with positive acceleration could be slowing down.
Explanation:
hope this helps
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There are two main reasons why a bouncing ball rise to a lower height with each bounce: 1) because every time it hits the ground it encounters friction, which slows it down, and every time it rises from the ground it encounters air resistance.
It’s not arrested because if you look closer it’s plug walk I don’t even understand how it talk tho
Density’s is mass by volume
So mass is 48 grams
Volume is 8cm^3
There is no need to convert units as they are same
So 48/8 gives you 6
So your answe should be 6g/cm^3 or 6gcm^-3
