A projectile is fired from ground level with a speed of 40 m/s at an angle of 30 degrees and hits a building 100m away.
What is the time of flight of the projectile?
How high above ground does the projectile hit the building?
What is the speed of the projectile just before it strikes the building?
Vo = 40 m/s @ 30 Deg.
Xo = 40*cos30 = 34.6 m/s.
Yo = 40*sin30 = 20 m/s.
a. Tr = (Yf-Yo)/g,
Tr = (0-20) / -9.8 = 2.04 s. = Time to
rise to max ht.
Tf = Tr = 2.04 s. = Fall time.
Tr + Tf = 2.04 + 2.04 = 4.08 s. = Time of flight.
b. hmax = Yo*t + 0.5g*t^2,
hmax = 20*2.04 - 4.9(2.04)^2=20.4 m.
Yf = Yo + gt,
Yf = 0 + 9.8*2.04 = 20 m/s.= Final velocity before striking gnd.
To find the time of flight, we can use the fact that the horizontal displacement of the projectile is 100m, and the initial velocity in the horizontal direction is 40 m/s. Using the equation:
Horizontal displacement = Initial velocity * Time of flight
We can rearrange this equation to solve for the time of flight:
Time of flight = Horizontal displacement / Initial velocity
Substituting the given values:
Time of flight = 100m / 40 m/s
Time of flight = 2.5 seconds
So, the time of flight of the projectile is 2.5 seconds.
To find the height above the ground where the projectile hits the building, we need to calculate the vertical distance traveled by the projectile. We can use the equation:
Vertical displacement = (Initial velocity * sin(θ))^2 / (2 * g)
Where θ is the launch angle (30 degrees) and g is the acceleration due to gravity (9.8 m/s^2).
Substituting the given values:
Vertical displacement = (40 m/s * sin(30))^2 / (2 * 9.8 m/s^2)
Vertical displacement = (20 m/s)^2 / 19.6 m/s^2
Vertical displacement = 400 m^2/s^2 / 19.6 m/s^2
Vertical displacement = 20.408 m
So, the projectile hits the building at a height of approximately 20.408 meters above the ground.
Finally, to find the speed of the projectile just before it strikes the building, we can use the equation of motion for the vertical component of velocity:
Final velocity = Initial velocity + (Acceleration * Time)
As the projectile reaches the building, its vertical velocity is equal to zero.
Final velocity = 0 m/s
Initial velocity in the vertical direction = Initial velocity * sin(θ) = 40 m/s * sin(30) = 40 m/s * 0.5 = 20 m/s
Acceleration = -g = -9.8 m/s^2
Using these values, we can rearrange the equation to solve for time:
Time = (Final velocity - Initial velocity) / Acceleration
Substituting the given values:
Time = (0 - 20 m/s) / (-9.8 m/s^2)
Time = -20 m/s / -9.8 m/s^2
Time = 2.04 seconds
So, the projectile takes approximately 2.04 seconds to reach the building.
The speed of the projectile just before it strikes the building will be equal to its initial vertical velocity, since there is no vertical acceleration acting on it.
Therefore, the speed of the projectile just before it strikes the building is 20 m/s.