Question

A projectile is launched upward at 24° from the ground and lands 12.5 s later at the same height. Assuming no air resistance...

What is the magnitude of the launch velocity?
How far from its original position did it land?

Answer 1

The magnitude of the launch velocity is 149.3 m/s

The distance is 74.65 m

Step-by-step explanation:

Given that,

Angle = 24°

Time =12.5 sec

We need to calculate the magnitude of the launch velocity

Using equation of motion

`y=-(1)/(2)gt^2+ut`

`y=-(1)/(2)gt^2+u\sin\theta* t`

Put the value in to the formula

`0=-(1)/(2)*9.8*(12.5)^2+u\sin24*12.5`

`u=(765.625)/(0.41*12.5)`

`u=149.3\ m/s`

We need to calculate the distance

Using formula of distance

`x=u\cos\theta* t`

Put the value into the formula

`x=149.3\cos24*12.5`

`x=74.65\ m`

Hence, The magnitude of the launch velocity is 149.3 m/s

The distance is 74.65 m