Answer to Question #176819 in Mechanics | Relativity for Felix

Question

A particle is projected from a point O with an initial velocity of 56m/s at an angle 60 to the horizontal. Find it's vertical displacement when its horizontal displacement is 70m

Answer 1

The displacement of the particle depends on time as

$x(t)=v_0 \cos{\alpha} t$

$\displaystyle y(t) = v_0 \sin \alpha \,t - \frac{gt^2}{2}$

$v_0 = 56\; m/s, \alpha =60^\circ, x =70\; m$

From the first equation:

$\displaystyle t = \frac{x}{v_0 \cos \alpha} = \frac{70}{56 \cdot 0.5} = 2.5 \; s$

Let's put this into the second equation:

$\displaystyle y(2.5 s) = 56 \cdot \frac{\sqrt3}{2} \cdot 2.5 - \frac{9.8 \cdot 2.5^2}{2} = 121.24 - 30.6= 90.61 \, m$

Answer: 90.61 m