Question

A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m vertically above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?​

Answer 1

Explanation:

Given:

t = 20 seconds

x = 3000 m

y = 450 m

a) To find the vertical component of the initial velocity $v_{0y}$, we can use the equation

$y = v_{0y}t - \frac{1}{2}gt^2$

Solving for $v_{0y}$,

$v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}$

$\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}$

$\:\:\:\:\:\:\:=120.5\:\text{m/s}$

b) We can solve for the horizontal component of the velocity $v_{0x}$ as

$x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}$

or

$v_{0x} = 150\:\text{m/s}$