Question

projectile is fired with an initial velocity of 193.2 feet per second upward at angle of 300to the horizontal from point 257.6 feet above a level plain. What horizontal distance will it cover before it strikes the level plain? A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.2 s later. What is the distance PQ? How far away from the dart board is the dart released? 3 Find the angle of elevation of a gun which fires a shell with muzzle velocity 366 m/s at a target on the same level but 4,575 m distant.

Answer 1

Explanations & Calculations

• Gravitational acceleration $\small \approx 9.8\,ms^{-2}\approx32.17\,ms^{-2}$

a)

• Apply suitable formula out of the 4 for both horizontal and vertical motions of the projectile.

$\qquad\qquad \begin{aligned} \small \uparrow\\ \small -257.6\,ft&=\small (193.2\sin30)\,fts^{-1}\times t+\frac{1}{2}\times(-32.17\,fts^{-2})\times t^2\\ \small t&=\small 8.0\quad or\quad-2.0\\ \small t&=\small 8.0\,s \end{aligned}$

• Then the horizontal coverage would be

$\qquad\qquad \begin{aligned} \small \to\\ \small s&=\small ut\\ &=\small 193.2.\cos30\,fts^{-1}\times8.0\,s \\ &=\small 1338.5\,ft \end{aligned}$

b)

• Apply $\small s= ut+1/2at^2$ downwards for the travel for that time.

$\qquad\qquad \begin{aligned} \small \downarrow\\ \small d&=\small 0+\frac{1}{2}\times(+9.8\,ms^{-2})\times(0.2\,s)^2\\ &\approx\small 0.2\,m \end{aligned}$

• Apply the same formula for the dart's horizontal motion.

$\qquad\qquad \begin{aligned} \small \to\\ \small &=\small 10\,ms^{-1}\times0.2\,s\\ &=\small 2\,m \end{aligned}$

c)

• Apply the standard formula for the horizontal range of a projectile for this case.

$\qquad\qquad \begin{aligned} \small R&=\small \frac{v^2}{g}.\sin2\theta\\ \small 4575\,m&=\small \frac{(366\,ms^{-1})^2}{9.8\,ms^{-2}}.\sin2\theta\\ \small \sin2\theta&=\small 0.33\\ \theta &=\small 9.6^0\\ \because\,\small\sin2\theta&=\small\cos(90-2\theta)\\ \theta &=\small90-9.6=80.4^0\\ \therefore\,\small \theta&=\small 9.6^0\quad or\quad 80.4^0 \end{aligned}$