A projectile, fired with unknown initial velocity, lands 18.4 s later on the side of a hill, 2740 m away horizontally and 469 m vertically above its starting point. (Ignore any effects due to air resistance.)
(a) What is the vertical component of its initial velocity?
115.741 m/s
(b) What is the horizontal component of its initial velocity?
148.91 m/s
(c) What was its maximum height above its launch point?
682.773 m
(d) As it hit the hill, what speed did it have and what angle did its velocity make with the vertical?
v = 162.384 m/s
θ = {I need the theta} °
To solve the problem, we can use the equations of motion for projectile motion. Before we get into the solution, let's break down the problem and understand the steps involved:
Given:
- Time of flight: 18.4 s
- Horizontal displacement: 2740 m
- Vertical displacement: 469 m
- No air resistance
We are asked to find:
(a) The vertical component of the initial velocity.
(b) The horizontal component of the initial velocity.
(c) The maximum height above the launch point.
(d) The speed and angle of the velocity just before hitting the hill.
Now let's go through each step to find the solutions:
(a) To find the vertical component of the initial velocity, we can use the equation:
Δy = Viy * t + (1/2) * g * t^2
Where:
Δy is the vertical displacement (469 m),
Viy is the vertical component of the initial velocity (what we need to find),
t is the time of flight (18.4 s),
g is the acceleration due to gravity (-9.8 m/s^2).
Rearranging the equation, we get:
Viy = (Δy - (1/2) * g * t^2) / t
Substituting the given values, we can calculate the vertical component of the initial velocity.
(b) To find the horizontal component of the initial velocity, we can use the equation:
Δx = Vix * t
Where:
Δx is the horizontal displacement (2740 m),
Vix is the horizontal component of the initial velocity (what we need to find),
t is the time of flight (18.4 s).
Rearranging the equation, we get:
Vix = Δx / t
Substituting the given values, we can calculate the horizontal component of the initial velocity.
(c) To find the maximum height above the launch point, we can use the equation for vertical displacement at the peak:
Δy = (Viy^2) / (2 * |g|)
Substituting the value of Viy calculated in part (a) and the value of g, we can find the maximum height.
(d) To find the speed just before hitting the hill and the angle it makes with the vertical, we need to consider the horizontal and vertical components of velocity. Since there is no air resistance, the speed remains constant throughout the flight.
The speed can be calculated as:
v = sqrt((Vix^2) + (Viy^2))
And the angle (θ) can be calculated as:
θ = arctan(Viy / Vix)
Substituting the values, we can find the speed and angle of velocity just before hitting the hill.
By following these steps and performing the necessary calculations, you can find the answers to the given questions.