A cannonball is shot from the top of a 30
m high hill at a velocity of 25 m/s and at
an angle of 40° above the horizontal and
lands on the ground below. The speed of
the cannonball at the top of its path is
most nearly
(A) 0
(B) 16 m/s
(C) 19 m/s
(D) 25 m/s
(E) 29 m/s

C) 19 m/s

Explanation:

The motion of the cannonball is a projectile motion, which consists of 2 independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

As a result, we have the following:

- The horizontal velocity of the cannonball remains constant during the motion, and it is given by

where

u = 25 m/s is the initial velocity

is the angle

Substituting,

- The vertical velocity keeps changing during the motion due to the acceleration of gravity. However, at the top of the trajectory, the vertical velocity is zero:

This means that at the top of its path, the cannonball has only horizontal velocity, so its velocity is

C) 19 m/s

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day