Question

If a projectile was shot from a building with a initial velocity of `60m/s` at an angle of`30^o` and took 3 sec to reach the ground,find the height of the building?

Answer 1

The height reached above the building is 46 m.

Explanation:

Use:

`sf(s=ut+1/2"at^2)`

This becomes:

`sf(h=vsinthetat-1/2"g"t^2)`

`sf(h=(60xx0.5xx3)-0.5xx9.81xx9color(white)(x)m)`

`sf(h=90.0-44.5=45.9color(white)(x)m)`

The total time of flight must be incorrect. If you read the comments 3 seconds must refer to the time to reach the zenith. This means the height reached above the building is 46 m.