Question

An arrow is shot at a 30 degree angle with the horizontal. it has a velocity of 49 m/s how high will the arrow go? what horizontal distance will the arrow travel? and how long will the arrow be in the air?

Answer 1

That arrow goes 30.59 m high, it travels 211.96 m horizontally and taken 4.99 seconds to land.

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at  

Initial velocity, u = u sin θ

Final velocity, v = -u sin θ  

Acceleration = -g

Substituting  

v = u + at  

-u sin θ   = u sin θ - g t

$t=\frac{2usin\theta }{g}$

This is the time of flight.

Consider the vertical motion of ball till maximum height,

We have equation of motion v² = u² + 2as

Initial velocity, u = u sin θ

Acceleration, a = -g  

Final velocity, v = 0 m/s  

Substituting  

v² = u² + 2as  

0² = u²sin² θ + 2 x -g x H

$H=\frac{u^2sin^2\theta }{2g}$

This is the maximum height reached,  

Consider the horizontal motion of ball,

Initial velocity, u = u cos θ

Acceleration, a =0 m/s²  

Time, $t=\frac{usin\theta }{g}$  

Substituting  

s = ut + 0.5 at²  

$s=ucos\theta \times \frac{2usin\theta }{g}+0.5\times 0\times (\frac{2usin\theta }{g})^2\\\\s=\frac{2u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{g}$

This is the range.

Here u = 49 m/s and θ = 30 degrees

Substituting

$t=\frac{2usin\theta }{g}=\frac{2\times 49\times sin30 }{9.81}=4.99s\\\\H=\frac{u^2sin^2\theta }{2g}=\frac{49^2\times sin^230 }{2\times 9.81}=30.59m\\\\R=\frac{u^2sin2\theta}{g}=\frac{49^2\times sin(2\times 30)}{9.81}=211.96m$

That arrow goes 30.59 m high, it travels 211.96 m horizontally and taken 4.99 seconds to land.