Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with speed u at an angle 'theta' with the horizontal?

Question

Answer

Problem:

Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with speed u at an angle 'theta' with the horizontal?

Solution:

Step 1: Find maximum height

The maximum height of the projectile can be found using the formula:

H = (u^2*sin^2(theta))/(2g)

Step 2: Find the time taken to reach half the maximum height

The time taken to reach half the maximum height can be found using the formula:

t = (u*sin(theta))/g

Let T be the total time taken for the projectile to reach the maximum height and come back to the ground. Therefore, the time taken to reach half the maximum height is T/2.

Step 3: Find the horizontal displacement at half the maximum height

The horizontal displacement at half the maximum height can be found using the formula:

X = (u^2*sin(theta)*cos(theta))/g

Step 4: Find the vertical velocity at half the maximum height

The vertical velocity at half the maximum height can be found using the formula:

V = u*sin(theta)/2

Step 5: Find the average velocity

The average velocity can be found using the formula:

Vavg = X/(T/2)

Therefore, the average velocity of the projectile between the instants it crosses half the maximum height is:

Vavg = (u^2*sin(theta)*cos(theta))/g*T

where T = 2*(u*sin(theta))/g is the total time taken for the projectile to reach the maximum height and come back to the ground.

Answer

Average velocity=displacement/time taken displacement=h/2,time taken can be find ,. h/2=u sin theta -1/2gt2 k,kinematic equation

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