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Lostsunrise [7]
3 years ago
11

A model rocket is launched at an angle of 70° above the x axis, with an initial velocity of 40 m/s. How high will the rocket be

5 seconds after launch?
A. 36.8 m
B. 65.4 m
C. 84.9 m
D. 122.5 m
Physics
2 answers:
s2008m [1.1K]3 years ago
7 0

Answer:

65.4 is your answer

Explanation:

Ilia_Sergeevich [38]3 years ago
4 0

Option (B) is correct.

using kinematic equation

h= V sin\theta t+ \frac{1}{2}gt^2

h= height of rocket

V= initial velocity= 40 m/s

t= time= 5 s

g= -9.8 m/s²

h=40 sin70 (5)+1/2 (-9.8)(5)²

h=65.4 m

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How do I calculate the tension in the horizontal string?
matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

T_1\cos 30\degree-m\cdot g=0

Solve for T₁,

T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N

Now, we use the second equation to find the tension in the horizontal string,

T_2-T_1\sin 30\degree=0

Solve for T₂,

T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0
1 year ago
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