A toy rocket is fired off a platform with an initial speed of 223 ft/sec at an angle of elevation of 32 degrees and achieves a maximum height of 291 ft. How how above the ground was the platform?
you would draw the picture and get the initial y velocity.
then, use any one of the possible equations to get the max height it goes. (at max height, velocity in the y direction is 0)
when you get that height, subtract it from the max to get the initial platform height
Can you give us the weight of the toy rocket
The distance Y that is rises above the platform is given by
M g Y = (1/2)M (Vsin32)^2
Y = [1/(2g)](118.2 ft/s)^2 = 217 ft
The above is a conservation of energy equation that uses only the vertical-motion part of the kinetic energy. The Vcos32 part of the KE does not change.
Since the toy rocket rises 291 ft from ground level, the platform must be 74 feet high.
I used 32.2 ft/s^2 for g
To find how high above the ground the platform was, we need to determine the maximum height reached by the rocket.
Given:
Initial speed (u) = 223 ft/sec
Angle of elevation (θ) = 32 degrees
Maximum height (h) = 291 ft
To find the maximum height (h), we can use the equations of motion in projectile motion. Let's break down the initial velocity into horizontal (u_x) and vertical (u_y) components.
Horizontal Component:
u_x = u * cos(θ)
Vertical Component:
u_y = u * sin(θ)
Using these components, we can calculate time of flight (t) using the vertical motion equation:
h = u_y * t + (1/2) * g * t^2
where g is the acceleration due to gravity (32 ft/sec^2). Since the rocket reaches its maximum height when its vertical component of velocity becomes zero, we can find the time of flight (t) using:
0 = u_y - g * t
Now, solving the equation:
t = u_y / g
Substituting the horizontal component (u_x) and vertical component (u_y) equations, we get:
t = (u * sin(θ)) / g
We can substitute this value of t in the equation for h to find the maximum height (h):
h = (u * sin(θ))^2 / (2 * g)
Plugging in the given values:
h = (223 * sin(32))^2 / (2 * 32)
Calculating this expression:
h ≈ 152.58 ft
Therefore, the platform was approximately 152.58 ft above the ground.