Question

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface An ideal projectile is launched from level ground at a launch angle of 26° and an initial speed of 48 m/sec. How far away from the launch point does the projectile hit the ground

Answer 1

Answer:

d = 185.26 meters

Explanation:

It is given that,

Launching angle of the projectile, $\theta=26^{\circ}$

Initial speed of the projectile, u = 48 m/s

Let at distance d the projectile hits the ground from the launch point. It is equal to range of the projectile. Its formula is given by :

$d=\dfrac{u^2\ sin2\theta}{g}$

Substituting all the values in above formula. So, we get :

$d=\dfrac{(48)^2\ sin2(26)}{9.8}$

d = 185.26 meters

So, the distance between the launch point and the point where it hits is 185.26 meters. Hence, this is the required solution.