Answer: i think its local drive
Explanation:
Consider horizontal component:
Using the formula:v2=u2+2a<span>s</span>
<span>0=(16sin<span>470</span><span>)2</span>+2(−9.81)s</span><span>s=6.98m</span><span>=7.0m(2s.f.)</span>The maximu height from the ground is 8.5m
Answer:
100 J
Explanation:
From the question, The work done by the forces in moving the box is given as
W = FxdcosФ+Fydcosα................... Equation 1
Where W = Work done, Fx = force acting parallel to the floor, d = distance moved by the box, Ф = angle the parallel force makes with the floor, Fy = force acting perpendicular to the floor, α = angle the perpendicular force make with the floor.
Give: Fx = 10 N, d = 20 m, Fy = 5 N, Ф = 0°, α = 90°
Substitute into equation 1
W = 10×10×cos0°+5×20×cos90°
W = 10×10×1+0
W = 100 J.
Note: The work done by the perpendicular force is zero
Hence the work done = 100 J
First we can say that since there is no external force on this system so momentum is always conserved.
now by the condition of elastic collision
![v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6](https://tex.z-dn.net/?f=v_%7B2f%7D%20-%20v_%7B1f%7D%20%3D%200.8%20-%200%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Enow%20add%20two%20equations%3C%2Fp%3E%3Cp%3E%5Btex%5D3%2Av_%7B2f%7D%20%3D%201.6)
also from above equation we have
So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.
