Answer:
Explanation:
Given
Initial speed is u=V
Maximum height of Pebble is H
Deriving maximum height of Pebble and considering motion in vertical direction
[tex]v^2-u^2=2 as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=Displacement
Final velocity will be zero at maximum height
[tex]0-(V)^2=2\times (-g)\cdot H[/tex]
[tex]H=\frac{V^2}{2g}[/tex]
i.e. maximum height is dependent on square of initial velocity
for twice the height
[tex]2H=\frac{(V')^2}{2g}[/tex]
on comparing
[tex]V'=\sqrt{2}V[/tex]
The speed needed for the pebble to go twice as high is a factor of √2
Newton's law of motion is given by:
v² = u² + 2gH
Where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and H is the height.
At maximum height, v = 0, g is negative(going up), hence:
0 = u² - 2gH
H = u²/2g
u = √(2gH)
To go twice as high:
u' = √(2g*2H) = u√2
The speed needed for the pebble to go twice as high is a factor of √2
Find out more at: https://brainly.com/question/14368172
