consider the motion of projectile A in vertical direction :
v₀ = initial velocity of projectile A in vertical direction = 0 m/s (since the projectile was launched horizontally)
a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²
t = time of travel for projectile A = 3.0 seconds
Y = vertical displacement of projectile A = height of the cliff = h = ?
using the kinematics equation along the vertical direction as
Y = v₀ t + (0.5) a t²
h = (0) (3.0) + (0.5) (9.8) (3.0)²
h = 44.1 m
Answer: 
Explanation:
Given
mass of ball m=10 kg
It is placed at a height of 150 m
It is dropped from the height and allowed to free fall for 40 m
Velocity acquired by the ball during this fall is given by 
Insert u=0, a=g
Kinetic energy at this instant
Answer:
Option B is the correct answer.
Explanation:
Let us consider 40 meter above ground as origin.
Initial velocity = 17 m/s
Final velocity = 24 m/s
Acceleration = 9.81 m/s
We have equation of motion v² = u² + 2as
Substituting
24² = 17² + 2 x 9.81 x s
s = 14.63 m
Distance traveled by rock = 14.63 m down.
Height of rock from ground = 40 - 14.63 = 25.37 m = 25.4 m
Option B is the correct answer.
The answer to the question is friction
