Question

A man, located a distance d=3.0 m from a target, fires a projectile at an angle θ=70 ​∘ ​​ above the horizontal. If the initial speed of the projectile is v=39.4 m/s, at what height hh does the projectile strike the building?

Answer 1

Answer:

So at height of 139.874 m projectile will strike the building              

Explanation:

We have given that man is located at a distance of 3 m

Angle of projection $\Theta =70^{\circ}$

Initial velocity is given as $u=39.4m/sec$

Height of projectile motion is given by

$h=\frac{u^2sin^2{\Theta }}{g}$

So $h=\frac{39.4^2sin^2{70^{\circ} }}{9.8}=139.874m$

So at height of 139.874 m projectile will strike the building