Question

A projectile is shot at an angle of `pi/12` and a velocity of `4 m/s`. How far away will the projectile land?

Answer 1

Answer is:

`s=0.8m`

Explanation:

Let the gravity acceleration be `g=10m/s^2`

The time traveled will be equal to the time it reaches its maximum height `t_1` plus the time it hits the ground `t_2`. These two times can be calculated from its vertical motion:

The initial vertical speed is:

`u_y=u_0sinθ=4*sin(π/12)`
`u_y=1.035m/s`

Time to maximum height `t_1`

As the object decelerates:

`u=u_y-g*t_1`

Since the object finally stops `u=0`

`0=1.035-10t_1`

`t_1=1.035/10`

`t_1=0.1035s`

Time to hit the ground `t_2`

The height during the rising time was:

`h=u_y*t_1-1/2*g*t_1^2`

`h=1.035*0.1035-1/2*10*0.1035^2`

`h=0.05359m`

The same height applies to the dropping time, but with the free fall formula:

`h=1/2*g*t_2^2`

`t_2=sqrt((2h)/g)`

`t_2=0.1035s`

(Note: `t_1=t_2` because of energy preservation law.)

The total time traveled is:

`t_t=t_1+t_2`

`t_t=0.1035+0.1035`

`t_t=0.207s`

The distance traveled in the horizontal plane has a constant speed equal to:

`u_x=u_0cosθ=4*cos(π/12)`
`u_x=3.864m/s`

Finally, the distance is given:

`u_x=s/t`

`s=u_x*t`

`s=3.864*0.207`

`s=0.8m`

P.S. For future problems identical to this one but with different numbers, you can use the formula:

`s=u_0^2*sin(2θ)/g`

Proof : we are basically going to use the same method inversely, but without substituting the numbers:

`s=u_x*t_t`

`s=u_0cosθ*2t`

`s=u_0cosθ*2u_y/g`

`s=u_0cosθ*2(u_0sinθ)/g`

`s=u_0^2*(2sinθcosθ)*1/g`

`s=u_0^2*sin(2θ)*1/g`

`s=u_0^2*sin(2θ)/g`