Question

A projectile is shot at an angle of `pi/6` and a velocity of `5 m/s`. How far away will the projectile land?

Answer 1

2.2 m

Explanation:

To solve this kinematics problem, we have to consider motion in both the horizontal and vertical component. We then realise that the t value (time of flight) would be the same for both components.

Vertical component:

Given `u=5sin(pi/6)=2.5m/s`,
`v=-u=-2.5m/s`
and `a=-9.81m/(s^2)` ,
find t.

`v=u+at`
`t=(v-u)/a=(-2.5-2.5)/(-9.81)=0.5097s`

Horizontal component:

Given `t=0.5097s`
and `v=5cos(pi/6)=4.330m/s` ,
find d.

distance
= speed x time
= (4.330)(0.5097)
= 2.207 m

Thus, the projectile will land 2.2 m away.