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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

User Teyzer
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4 votes

Answer:

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.

User Koehlma
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