Answer to Question #163095 in Mechanics | Relativity for Akinola

Question

A projectile motion us given by y=√3/3x - 4/3x^2 where (x,y) is the coordinate of the projectile at any time t seconds. Both x&y are measured in meters.

a) find the range of the projectile

b) will the projectile hit the too of the cliff 50m high?

c) what is the coordinate of the projectile at t=6seconds

Answer 1

Solution:

x = v₀x t

y = v₀y t - $\frac{gt^2}{2}$

eliminating t on get the trajectory:

$y=tan(\alpha)x-[\frac{g}{2v_0x^2}]x^2$

then

$tan(\alpha)= \sqrt{\smash[b]{\frac{3}{3}}} \rightarrow\; \alpha=30 ^o$

$\dfrac{g}{2v_ox^2}=\dfrac{4}{3}\rightarrow v_ox=1.92(\frac{m}{s})\rightarrow v_o=2.21(\frac{m}{s})$

so:

a)

$x=\dfrac{v_o^2sin(2 \alpha)}{g}=0.43m$

b)

$h=\dfrac{v_oy^2}{2g}=0.06m$

c)

fly time t_f is

t_f = $\dfrac{2v_oy}{g}=0.23s$

then being t = 6 s > t_f

x = v_ox t_f = 0.43 m

y = 0