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marta [7]
3 years ago
11

Solve for angles A-D

Physics
1 answer:
patriot [66]3 years ago
7 0
B = 53°

Reason: Opposite angles

A = 37°

Reason: Sum of angels on a straight line = 180 so 180-90-53 = 37°
C = 90°

Reason: Right angles = 90°

D = 37°

Reason: Sum of angels on a straight line = 180 so 180-90-53 = 37°
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A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert
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The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

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Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

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So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

F_y=F-mg=8.5-0.46\times 9.8=3.992\ N

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2

Vertical displacement of rocket is same as the height of loop. So, y=20\ m

There is no initial velocity in the vertical direction. So, u_y=0\ m/s

Now, applying equation of motion in vertical direction. we have:

y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

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