Question

A projectile is fired with an initial velocity of 60 m/sec. Upward at an angle of 30° to the horizontal from a point 80m above a level plain. What horizontal distance will it cover before it strikes the plain?

Answer 1

Let x-axis be directed horizontally from the point of the start of the motion and y-axis be directed vertically from the same point.

The projection of the velocity on the x-axis is $v_x = v\cos\alpha,$ the projection on the y-axis is $v_y = v\sin\alpha.$ The x-projection is the same all the time of motion and y-projection is $v_y = v\sin\alpha - gt.$

The coordinates after time t are $x(t) = v\cos\alpha\cdot t,$ $y(t) = v\sin\alpha\cdot t - \dfrac{gt^2}{2}.$

Let us determine the time t when y = -80

$60\cdot \sin30^\circ t - 5t^2 = -80.\\ t_1 = -2, \;\; t_2 = 8.$

We are interested in positive value, so t = 8 seconds.

The x-coordinate is $60\cos30^\circ\cdot 8 = 416$ m.