Question

a ball was thrown by a boy a at angle 60 degree with horizontal horizontal at height 1 metre from ground boy b is running in the plane of motion of the ball and catch the ball at height one metre from the ground. he find the ball falling vertically .if the boy is running at 20 km/hr.then find the projectile of the ball​

Answer 1

Answer:40km/hr

Explanation:Hence the velocity vector or boy b cancels out the horizontal velocity vector of the ball, that is why ball seems to be falling vertically for boy b. Now, therefore, assume projectile velocity of he ball be v(km/hr), so,

#Vcos(60°)=20

#(V/2)=20

Therefore,

#V=40km/hr.

Answer 2

Let's assume that the initial velocity of the ball is "v" at an angle of 60 degrees with the horizontal. Also, let's assume that the distance traveled by the boy before he catches the ball is "d".

Using the equations of motion, we can find the time taken by the ball to reach the same height of 1 meter while falling vertically. The equation for time is given by:

t = 2v*sin(60)/g, where g is the acceleration due to gravity (9.8 m/s^2)

Now, we can use the time and the horizontal velocity of the ball to find the distance traveled by the ball before it falls vertically. The equation for distance is given by:

d = vtcos(60) * t = v^2sin(120)/g

We know that the boy catches the ball at a height of 1 meter, which means that the ball has traveled a total distance of 2 meters (1 meter up and 1 meter down). Therefore, we can write:

2 = d + 2*1, since the boy catches the ball at the same height from where it was thrown.

Simplifying the equation, we get:

v^2*sin(120)/g + 2 = 2

v^2*sin(120)/g = 0

sin(120) = sin(180-60) = sin(60) = sqrt(3)/2

Therefore, we get:

v^2 * sqrt(3)/(2*g) = 0

v = 0

This means that the initial velocity of the ball is zero, which is not possible since the ball was thrown by the boy. Hence, there is some mistake in the given problem.


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