Angular relationship question.

[CandyApples]

Homework Statement

A model rocket is launched at 50m/s, 35 degrees above horizontal. What is the horizontal displacement when its velocity vector is at 25 degrees.

Homework Equations

Kinematic equations, the relationship between sin cos and tan.

The Attempt at a Solution

tan(25)= .4663. I interpret this as the y (sin) velocity being .4663 the x velocity. Vy = .4663Vx

Vy = 50sin(25)-9.8t
.4663Vx = 50sin(25)-9.8t
Vx = (50sin25-9.8t)/.4663
Vx also equals 50*cos25 therefore...
50cos25 = (50sin25-9.8t)/.4663
(.4663*50cos25-50sin25)/-9.8 = t
t = 3.54E^-5.

Something about this answer seems very very wrong, especially when plugged back into the x displacement equation.

 

[ideasrule]

Vx also equals 50*cos25 therefore...

Why 50? The rocket decelerated due to gravity.

 

[tomcue]

I would approach this problem by first drawing a diagram to visualize the situation. From the given information, we know that the rocket is launched at a velocity of 50m/s at an angle of 35 degrees above the horizontal. We are then asked to find the horizontal displacement when the velocity vector is at an angle of 25 degrees.

To solve this problem, we can use the kinematic equations to find the horizontal and vertical components of the rocket's velocity at the given angle. We can then use these components to calculate the displacement at the given time.

Based on the given information, we can use the following equations:

Horizontal component of velocity (Vx) = Vcosθ
Vertical component of velocity (Vy) = Vsinθ
Horizontal displacement (x) = Vx * t
Vertical displacement (y) = Vy * t - 1/2 * g * t^2

Using these equations, we can calculate the horizontal displacement when the velocity vector is at an angle of 25 degrees. Plugging in the given values, we get:

Vx = 50cos35 = 40.82 m/s
Vy = 50sin35 = 28.79 m/s
x = Vx * t = 40.82 * t
y = Vy * t - 1/2 * g * t^2 = 28.79 * t - 4.9 * t^2

To find the time (t) when the velocity vector is at an angle of 25 degrees, we can use the inverse tangent function:

tan25 = y/x
t = arctan(y/x) = arctan(28.79/40.82) = 0.62 seconds

Plugging this value of t back into the horizontal displacement equation, we get:

x = 40.82 * 0.62 = 25.36 meters

Therefore, the horizontal displacement when the velocity vector is at an angle of 25 degrees is approximately 25.36 meters.

In conclusion, as a scientist, I would approach this problem by breaking it down into smaller components, using relevant equations, and double-checking my calculations to ensure accuracy. I would also make sure to include units in my calculations and final answer.