A projectile, fired with unknown initial velocity, lands 20.8 s later on the side of a hill, 2900 m away horizontally and 411 m vertically above its starting point. (Ignore any effects due to air resistance.)
(a) What is the vertical component of its initial velocity?
(b) What is the horizontal component of its initial velocity?
(c) What was its maximum height above its launch point?
(d) As it hit the hill, what speed did it have and what angle did its velocity make with the vertical?
To find the answers to these questions, we can use the equations of motion for projectile motion. Let's break it down step by step:
Step 1: Find the time of flight (total time taken) for the projectile.
Given: Total time of flight (t) = 20.8 seconds.
Step 2: Find the vertical component of the initial velocity.
Given: Vertical displacement (y) = 411 m.
The equation for vertical displacement in projectile motion is:
y = (viy * t) + (1/2 * g * t^2)
Where viy is the initial vertical velocity and g is the acceleration due to gravity (approximately 9.8 m/s^2). Substituting the known values into the equation:
411 = (viy * 20.8) + (1/2 * 9.8 * (20.8)^2)
This equation can be rearranged to solve for viy:
viy = (y - (1/2 * g * t^2)) / t
Step 3: Find the horizontal component of the initial velocity.
Given: Horizontal displacement (x) = 2900 m and t = 20.8 seconds.
The equation for horizontal displacement in projectile motion is:
x = vix * t
Where vix is the initial horizontal velocity.
Solving for vix:
vix = x / t
Step 4: Find the maximum height above the launch point.
At the maximum height, the vertical velocity (viy) becomes zero. Using the equation:
viy = viy0 - g * t
We can find the time taken to reach the maximum height.
t = viy0 / g
Substituting the known values:
t = viy / g
The maximum height (ymax) can be found using the equation:
ymax = viy0 * t - (1/2 * g * t^2)
Step 5: Find the speed at the moment it hits the hill and the angle it makes with the vertical.
Given: Vertical and horizontal components of the velocity (viy and vix).
The speed can be found using the equation for total velocity:
v = √(vix^2 + viy^2)
The angle θ that the velocity makes with the vertical can be calculated using the inverse tangent:
θ = tan^(-1)(vix / viy)
Now, let's plug in the values and calculate the answers:
(a) Vertical component of initial velocity (viy):
Substituting the known values into the equation for viy:
viy = (y - (1/2 * g * t^2)) / t
(b) Horizontal component of initial velocity (vix):
Substituting the known values into the equation for vix:
vix = x / t
(c) Maximum height above the launch point (ymax):
Substituting the known values into the equation for ymax:
ymax = viy * t - (1/2 * g * t^2)
(d) Speed at the moment it hits the hill and the angle it makes with the vertical (v and θ):
Substituting the known values into the equations for v and θ:
v = √(vix^2 + viy^2)
θ = tan^(-1)(vix / viy)
Using these steps, you can find the answers to each part of the problem.