A 2 kg ball rolls off a 27 m high cliff, and lands 25 m from the base of the cliff. Express the displacement and the gravitational force in terms of vectors and calculate the work done by the gravitational force. Note that the gravitational force is < 0, -mg, 0 >, where g is a positive number ( 9.8 N/kg). (Let the origin be at the base of the cliff, with the x direction towards where the ball lands, and the y direction taken to be upwards.)

r = < 10.6t, 27 – 4.9t², 0 >m

F < 0, -19.6, 0 > N

Explanation:

Given the mass of the ball

m = 2.0kg , g = 9.8m/s²

H = 27m Range = x = 25m

X = Xo + Vo×t×Cosθ

y = yo + Vo×t×Sinθ -1/2gt²

But θ = 0° the ball rolls off the cliff

y = 27 + Vot ×0 – 1/2gt²

y = 27–1/2gt²

X = Xo + Vo×t×Cos0°

X = Xo + Vot

Taking the base of the cliff as Xo = 0m

X = 0 + Vot =Vot

We need to calculate the velocity Vo the initial velocity in order to find expressions for the displacement along the given axes and to do this we need to know how long this motion lasted for, t.

So to calculate t, we need to find the time the stone strikes the ground. y = 0m

So,

0 = 27 – 1/2gt²

1/2gt² = 27

t² = 27×2/g = 54/9.8

t² = 5.51

t = 2.35s

The ball travels a horizontal distance of 25m from the base of the cliff. So

X = 25 = Vo ×2.35

Vo = 25/2.35 = 10.6m/s

X = 10.6t (substituting for Vo)

y = 27 – 1/2×9.8t² = 27 – 4.9t²

Fy = -mg = -2×9.8 = ‐19.6N

So the expression for the displacement

r = < 10.6t, 27 – 4.9t², 0 >m

F < 0, -19.6, 0 > N

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle