B. velocity at position x, velocity at position x=0, position x, and the original position
In the equation
=
+2 a x (x - x₀)
= velocity at position "x"
= velocity at position "x = 0 "
x = final position
= initial position of the object at the start of the motion
1)
Answer:
Part 1)
H = 30.6 m
Part 2)
t = 2.5 s
Part 3)
t = 2.5 s
Part 4)

Explanation:
Part 1)
initial speed of the ball upwards

so maximum height of the ball is given by



Part 2)
As we know that final speed will be zero at maximum height
so we will have



Part 3)
Since the time of ascent of ball is same as time of decent of the ball
so here ball will same time to hit the ground back
so here it is given as
t = 2.5 s
Part 4)
since the acceleration due to earth will be same during its return path as well as the time of the motion is also same
so here its final speed will be same as that of initial speed
so we have

2)
Answer:
a = 9.76 m/s/s
Explanation:
As we know that the object is released from rest
so the displacement of the object in vertical direction is given as



3)
Answer:
v = 29.7 m/s
Explanation:
acceleration of the rocket is given as

time taken by the rocket
t = 0.33 min
final speed of the rocket is given as



4)
Answer:
Part 1)
y = 25.95 m
Part 2)
d = 6.72 m
Explanation:
Part 1)
As it took t = 2.3 s to hit the water surface
so here we will have



Part 2)
Distance traveled by it in horizontal direction is given as



P = m x v
P = 30 x 10
=300
As we know that the formula of kinetic energy will be

now here we know that
m = 2 kg
v = 1 m/s
so from the above equation we have

