Vertical Projectile Motion: Height and Time Calculations

Question:

A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a multistorey building . The height of the point from where the ball is thrown is 25.0 m from the ground.(a) How high will the ball rise? and (b) How long will it be before the ball hits the ground?

Solution: Height and time calculation

a) How high will the ball rise?

Consider the vertically upward direction with zero at the ground.

Initial velocity u =+20 ms-1 (upward direction)

Final velocity v = 0 (maximum point the velocity is zero)

The ball rises to height h (displacement is positive)

Acceleration due to gravity a=-g

Applying the third equation of motion

$${v^2 = u^2 +2 a s}$$

$${0 = 20^2 +2 (-10) (s)}$$

Here s=y-y_o (Total height from the ground – height of the building)

$${400=20s}$$

$${s=20 m}$$

Here y-y_o = 20 m =s

The ball ill rise to 20 m from the top of the multistorey building.

(b) How long will it be before the ball hits the ground?

(i) Total time taken = Time taken from A to B +Time taken from B to C

Time taken from A to B (thrown up from the top of the building) then the equation is

$${u=gt}$$

(Refer the image of the sign convention)

$${t=\frac{u}{g}}$$

$${t=\frac{20} {10} =2 s}$$

(ii) Time taken from B to C ( dropped from the maximum height)

$${h=\frac{1} {2} g t^2}$$

$${t=\sqrt\frac{2h} {g}}$$

Height of the ball from the ground is 25 m +20 m = 45 m

$${t=\sqrt\frac {90}{10}}$$

$${t=\sqrt9}$$

$${t=3 s}$$

Total time taken = Time taken from A to B +Time taken from B to C

=2 s + 3 s = 5 s