Question

At what
аngle with respect to the horizontal, should a projectile be thrown with a velocity of 19.6m/s, to just clear a wall 14.7m high at a distance of 19.6m?​

Answer 1

Given:

Velocity = 19.6 m/s

Height of the wall = 14.7 m

Distance of the wall = 19.6 m

To find:

The angle at which the projectile should be thrown

Calculation:

By using the equation of trajectory

    $Y=xtan\theta-\frac{gx^2}{2u^2cos^2\theta}$  

By substituting the given values in the above equation

    $14.7=(19.6)tan\theta-\frac{9.8\times(19.6)^2}{2\times(19.6)^2\times cos^2\theta}$

    $14.7=(19.6)tan\theta-\frac{9.8\times(19.6)^2\times sec^2\theta}{2\times(19.6)^2}$

    $14.7=(19.6)tan\theta-4.9sec^2\theta$

    $14.7=(19.6)tan\theta-4.9(1+tan^2\theta)$

    $4.9tan^2\theta-19.6tan\theta+19.6=0$

    $tan\theta=2$

    $\theta=63.43^{\circ}$

Final answer:

The angle at which the projectile should be thrown Is 63.43°