From a top a building is 20 m tall, you launch a projectile up at an angle of 30 degrees to the horizontal. You launch at 80 m/s find a) the total time in flight, b) the hightest point reached above the building, and c) distance between the landing point and the base of the building.
I need help with the equations to get me started.
break the initial velocity into vertical and horizontal components.
Then, use the vertical component in the height equation to find the time of flight. Use the quadratic equation.
For the max height, use the vertical velocity equation (velocity at the top is zero), find the time, then put that into the height equation.
For the distance, use the horizontal distance equation
I will be happy to critique your work.
To solve this problem, we need to break down the initial velocity into its vertical and horizontal components.
The vertical component of the initial velocity can be found using the formula:
Viy = V * sin(θ)
Where:
Viy is the vertical component of the initial velocity
V is the initial velocity (80 m/s in this case)
θ is the launch angle (30 degrees in this case)
Plugging in the given values, we get:
Viy = 80 * sin(30)
Next, we can use the vertical component of the velocity to find the time of flight using the height equation:
y = Viy * t - (1/2) * g * t^2
Where:
y is the vertical displacement (20 m in this case, as the projectile starts from the top of the building and falls down)
t is the time of flight (we need to find this)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Rearranging the equation, we get:
(1/2) * g * t^2 - Viy * t + y = 0
This is a quadratic equation in terms of t. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = (1/2) * g, b = -Viy, and c = y. Plugging in the values, we can solve for t - the total time in flight.
To find the highest point reached above the building, we need to find the time at which the projectile reaches its maximum height. We can use the vertical velocity equation:
Vfy = Viy - g * t
At the highest point, the vertical velocity is zero. So we can set Vfy to zero and solve for t.
Lastly, to find the distance between the landing point and the base of the building, we can use the horizontal distance equation:
x = V * cos(θ) * t
Where:
x is the horizontal distance
V is the initial velocity (80 m/s in this case)
θ is the launch angle (30 degrees in this case)
t is the time of flight (found in the previous step)
Plugging in the values, we can calculate the distance between the landing point and the base of the building.
Once you have solved these equations, you should have all the answers for a), b), and c). Let me know if you need further assistance or if you want me to check your work.