Answer:
The new height the ball will reach = (1/4) of the initial height it reached.
Explanation:
The energy stored in any spring material is given as (1/2)kx²
This energy is converted to potential energy, mgH, of the ball at its maximum height.
If the initial height reached is H
And the initial compression of the spring = x
So, mgH = (1/2)kx²
H = kx²/2mg
The new compression, x₁ = x/2
New energy of loaded spring = (1/2)kx₁²
And the new potential energy = mgH₁
mgH₁ = (1/2)kx₁²
But x₁ = x/2
mgH₁ = (1/2)k(x/2)² = kx²/8
H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s
Answer:
may be upside down alphabet :"T"
Explanation:
Impulse describes the change of momentum. Since we don't know the momentum of the soccer ball before the hit, this question is hard to answer. If you assume the momentum of the ball before the hit was p = 0, then the change in momentum is just Δp = Impulse = mv.
X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75
