Question

A projectile is fired with in vertically upward direction. Show that it will strike the ground with same
speed with which it was fired?

Answer 1

Answer:

v

e

is the escape velocity then

2

1

v

e

2

=

R

GMm

Now the projectile is fired with speed kv

e

.

Then loss in kinetic energy is gain in potential energy

⟹

2

1

m(kv

e

)

2

=GMm(

R

1

−

h

1

)

⟹k

2

(

R

GMm

)=GMm(

R

1

−

h

1

)

⟹h=

1−k

2

R

.

Explanation:

hope it helps you dear

Answer 2

Answer:

Let us consider the fired projectile of mass m with initial velocity u.

At any point of time, the acceleration of the projectile is g (acceleration due to gravity).

Applying the second law of motion,

F = ma = m(du/dt)

Since, F = 0 (as there is no external force acting on the projectile)

=> m(du/dt) = 0

Integrating,

u = u + 0

=> u = constant

Hence, the projectile will strike the ground with the same speed with which it was fired.