Question

A projectile is fired vertically from earth's surface with an initial speed of 7.3 km/s. neglecting air drag, how far above the surface of earth will it go?

Answer 1

We can solve the problem by using conservation of energy.

In fact, initially the projectile has only kinetic energy, which is given by

`K= (1)/(2)mv^2`
where m is the projectile's mass while
`v=7.3 km/s=7300 m/s` is its initial velocity.

At the point of maximum height, the speed of the projectile is zero, so it only has gravitational potential energy which is equal to

`U=mgh`
where g is the gravitational acceleration and h is the maximum height of the projectile.

Since the energy must be conserved, we can equalize K and U to find the value of h:

`(1)/(2)mv^2=mgh`

`h= (v^2)/(2g)= ((7300 m/s)^2)/(2 \cdot 9.81 m/s^2)=2.72 \cdot 10^6 m = 2720 km`