Question

A ball thrown by a boy in the street is caught by another boy on balcony 4m above the ground and 18m away after 2s. Calculate the initial velocity and the angle of projection.

Answer 1

Let us first resolute the velocity as ux for velocity along x axis due to which it travels 18m and uy which is vertical velocity due to which it travels 4m.

There is no air drag(assumption)

So,

ux=s/t

ux= 9m/s

There is gravitational accn along y axis, 

So,

a=-10m/s^2

By solving this eq ( s= uyt + ½ at^2)

We get uy=12m/s

We know that,

Resultant = rt(a^2+ b^2 +2abcos(theta))

So, substituting values,

u= 15m/s

(note: here we have resoluted the initial velocity as ux and uy and the resolutions are perpendicular, so cos(pi/2) = 0 

So,

u=rt(ux^2+uy^2))

Now , to find the angle,

We use the formula 

tan(alpha)=uysin(theta)/(ux+uycos(theta))

Where alpha is the angle between the horizontal axis( or x axis, here it is ux) and theta is the angle btw the adding vectors...

If you want angle of projection with respect to y axis then just substitute ux with uy and uy with ux in above eq.)

By solving we get,

tan(alpha)
= uy/ux

=12/9

=4/3 (because sin90 is 1 ans cos90 is 0)

So alpha = tan^-1(4/3)

This may be kept as final answer but if we further solve,

Angle of projection=42.96 degree.

Answer 2

The projection angle is approximately 43 degrees.

To calculate the initial velocity and angle of projection of a ball thrown by a boy in the street and caught by another boy on a balcony 4 meters above the ground and 18 meters away after 2 seconds.

We need to first find the velocity along the x-axis (ux) and the velocity along the y-axis (uy).

The velocity along the x-axis can be found using the formula

ux = s/t

where s is the distance traveled along the x-axis (18 meters) and t is the time taken (2 seconds).

This gives us ux = 9 m/s.

The velocity along the y-axis can be found using the equation of motion, with acceleration due to gravity (-10 m/s^2) and the distance traveled along the y-axis (4 meters).

This gives us uy = 12 m/s.

The resultant velocity (u) can then be found using the equation

u = √(ux^2 + uy^2).

In this case, u = 15 m/s. The angle of projection can be found using the formula

tan(α) = uy/ux,

Where α is the angle between the x-axis and the projection vector. Solving this equation gives us

tan(α) = 4/3 and α = tan^-1(4/3), which is approximately 43 degrees. This is the angle of projection.

For more such questions on angle of projection: https://brainly.in/question/51891174

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