2. a cannonball is fired at an angle 0 = 35° to the horizontal, with a velocity
of 85 ms 1.
a) what is the maximum vertical height reached by the cannonball?
b) what is the maximum horizontal range of the cannonball?

a) 120 m

b) 690 m

Explanation:

Given:

x₀ = 0 m

y₀ = 0 m

v₀ = 85 m/s

θ = 35°

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

a) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find y.

vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)

(0 m/s)² = (85 sin 35° m/s)² + 2 (-9.8 m/s²) (y − 0 m)

y ≈ 120 m

b) At the maximum range, the cannonball lands, so y = 0.

First, we need to find the time it takes to land.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (0 m) + (85 sin 35° m/s) t + ½ (-9.8 m/s²) t²

t = 0 s, 9.95 s

Now find the final horizontal position x:

x = x₀ + v₀ₓ t + ½ aₓ t²

x = (0 m) + (85 cos 35° m/s) (9.95 s) + ½ (0 m/s²) (9.95 s)²

x ≈ 690 m

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

the answer would be a