Question

g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower

Answer 1

The projectile strikes the tower at a height of 354.824 meters.

Step-by-step explanation:

The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:

Horizontal motion

`x = x_(o)+v_(o)\cdot t \cdot \cos \theta`

Vertical motion

`y = y_(o) + v_(o)\cdot t \cdot \sin \theta +(1)/(2) \cdot g \cdot t^(2)`

Where:

`x_(o)`,
`x` - Initial and current horizontal position, measured in meters.

`y_(o)`,
`y` - Initial and current vertical position, measured in meters.

`v_(o)` - Initial speed, measured in meters per second.

`g` - Gravitational acceleration, measured in meters per square second.

`t` - Time, measured in seconds.

The time spent for the projectile to strike the tower is obtained from first equation:

`t = (x-x_(o))/(v_(o)\cdot \cos \theta)`

If
`x = 600\,m`,
`x_(o) = 0\,m`,
`v_(o) = 120\,(m)/(s)` and
`\theta = (\pi)/(4)`, then:

`t = (600\,m-0\,m)/(\left(120\,(m)/(s) \right)\cdot \cos (\pi)/(4) )`

`t \approx 7.071\,s`

Now, the height at which the projectile strikes the tower is: (
`y_(o) = 0\,m`,
`t \approx 7.071\,s`,
`v_(o) = 120\,(m)/(s)` and
`g = -9.807\,(m)/(s^(2))`)

`y = 0\,m + \left(120\,(m)/(s) \right)\cdot (7.071\,s)\cdot \sin (\pi)/(4)+(1)/(2)\cdot \left(-9.807\,(m)/(s^(2)) \right) \cdot (7.071\,s)^(2)`

`y \approx 354.824\,m`

The projectile strikes the tower at a height of 354.824 meters.