Question

What is the horizontal range of a projectile fired vertically upward?

Answer 1

Explanation:

Correct option is

B

1−k2R

If ve is the escape velocity then

21ve2=RGMm

Now the projectile is fired with speed kve.

Then loss in kinetic energy is gain in potential energy

⟹21m(kve)2=GMm(R1−h1)

⟹k2(RGMm)=GMm(R1−h1)

⟹h=1−k2R.

Answer 2

Answer:

Range=0

Explanation:

R=vi² sin2 theeta/g

for vertically upward , theeta will be 90°

so,

R= vi² sin2(90)/g

R=vi² sin 180/g

since, sin180=0

so,

R=vi² (0)/g

R=0