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vivado [14]
3 years ago
5

A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m verticall

y above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?​
Physics
1 answer:
lorasvet [3.4K]3 years ago
4 0

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}

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