Aball is launched from a cliff with and initial velocity of 15 m/s at an angle of 40 above horizontal. if the ball lands 17.5 m away, determine the height of the cliff.

Explanation:

Evaluate in the x direction to find the time it takes to land:

x = x₀ + v₀ₓ t + ½ aₓ t²

17.5 = 0 + (15 cos 40) t + ½ (0) t²

t = 1.52

Evaluate in the y direction to find the initial height:

y = y₀ + v₀ᵧ t + ½ g t²

0 = h + (15 sin 40) (1.52) + ½ (-9.8) (1.52)²

h = -3.32

The cliff is actually 3.32 meters below where it lands.  Make sure you copied the problem correctly.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2