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olga nikolaevna [1]
3 years ago
14

A boy kicks a football from ground level. The ball takes 3 seconds to reach its maximum height. What is the angle of the initial

velocity with respect to the horizontal, if the ball lands 15 m away from the launching point?
Physics
1 answer:
ruslelena [56]3 years ago
4 0
<h2>The angle of the initial velocity with respect to the horizontal is 85.14°</h2>

Explanation:

Given that the ball takes 3 seconds to reach its maximum height.

Consider the vertical motion of ball till maximum height.

We have equation of motion v = u + at

     Acceleration, a = -9.81 m/s²

     Final velocity, v = 0 m/s    

     Time, t = 3 s

     Substituting

                      v = u + at  

                      0 = u + -9.81 x 3

                      u = 29.43 m/s

Initial vertical velocity is 29.43 m/s.

Now consider horizontal motion of ball.

Time of flight of ball = 2 x Time to reach maximum height = 2 x 3 = 6 s

Displacement = 15 m

We have equation of motion s = ut + 0.5 at²

        Displacement, s = 15 m

        Acceleration, a = 0 m/s²  

        Time, t = 6 s      

     Substituting

                      s = ut + 0.5 at²

                      15 = u x 6 + 0.5 x 0 x 6²

                      u = 2.5 m/s

Initial horizontal velocity is 2.5 m/s

Let r be the initial velocity and θ be the angle with horizontal

              Initial vertical velocity = rsinθ = 29.43 m/s

              Initial horizontal velocity = rcosθ = 2.5 m/s

Dividing

              \frac{rsin\theta }{rcos\theta }=\frac{29.43}{2.5}\\\\tan\theta=11.77\\\\\theta=85.14^0

The angle of the initial velocity with respect to the horizontal is 85.14°

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The answer to this would be ocean floor
4 0
3 years ago
11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle
Alja [10]

Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Explanation:

To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.

For M vector you obtain:

M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}

For N vector:

N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}

The resultant vector is the sum of the components of M and N:

F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}

The magnitude of the resultant vector is:

|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm

And the direction of the vector is:

\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°

hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

4 0
3 years ago
A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi
bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc    where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

rf = \sqrt{rfX^2+rfY^2} = 245.45km

And the angle measure from the y-axis is:

\alpha =atan(rfX/rfY) = 21.45\°

So the answer is 245.45km in a direction 21.45° west of north from city A

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Answer:

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Explanation:

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