Three balls which have equal masses are fired with equal speeds. One ball is fired up at an angle of 45 degrees, the other is fired up at an angle of 60 degrees, and the third ball is fired straight up at an angle of 90 degrees from the horizontal surface. Rank in order, from largest to smallest, their speeds Va, Vb, and Vc as they cross a dashed horizontal line approximately a couple meters away from the balls. Explain. (All three are fired with sufficient speed to reach the line.)

Let [tex]V_{a}[/tex], [tex]V_{b}[/tex], [tex]V_{c}[/tex] be the speed of the balls which was fired at an angle of 45°, 60°, 90° respectively from the horizontal surface, when the balls crosses the dashed horizontal line a few meters away from the place where it is fired initially

After firing the only force that is acting on the ball will be the force of gravity assuming that there is no air resistance, so the acceleration will be g in the downward direction

Using the below formula we can calculate the final velocity of the ball after travelling some distance

v² - u² = 2×a×s

where v is the final velocity of the ball

u is the initial velocity of the ball

s is the distance between the points where initial velocity is u and final velocity is v

a is the acceleration of the ball

From this we get

v² = u² + 2×a×s

Here in this case we are considering in vertical direction and in this direction for all balls a = -g as g is acted on the balls opposite to the direction of motion

s is same for all balls

Let u be the initial speed of all balls

speed is same for all balls but not initial velocity in the vertical direction

For the ball that is fired from an angle of 90°, initial velocity is u

For the ball that is fired from an angle of 60°, initial velocity is u×sin60° = u×(√3 ÷ 2)

For the ball that is fired from an angle of 45°, initial velocity is u ×sin45° = u×(1 ÷ √2)

As a and s in the formula are constant for all particles, the final velocity will depend on their initial velocities

We can observe that as angle with horizontal is increasing the initial velocity is also increasing

∴ Final velocity in the vertical direction of the ball fired from 90° > Final velocity in the vertical direction of the ball fired from 60° > Final velocity in the vertical direction of the ball fired from 45°

As there is no acceleration in horizontal direction initial velocity in horizontal direction = final velocity in horizontal direction

In horizontal direction

velocity of the ball fired from 90° is 0

velocity of the ball fired from 60° is u×cos60° = u÷2

velocity of the ball fired from 45° is u ×sin45° = u×(1 ÷ √2)

Speed = √((velocity in horizontal direction)² +(velocity in vertical direction)²)

For all balls we get speed² = u² - 2×g×s

∴ Final speed of all balls is same