Question

At t=0 a ball is projected vertically upward from a roof of a building of height 68.1 meters with a velocity of 42.4 m/s. The ball rises, reaches a maximum height and then falls, missing the roof and strikes the ground below. Calculate the total time the ball is in flight.

Answer 1

Answer:

9.86s

Explanation:

The total time ($t_{T}$) the ball is in flight is the sum of the time taken ($t_{H}$) to reach maximum height and the time taken ($t_{G}$) to strike the ground from maximum height. i.e

$t_{T}$ = $t_{H}$ + $t_{G}$

Calculate time taken to reach maximum height, $t_{H}$.

Using one of the equations of motion;

v = u + at   ------------------------(i)

Where;

v = final velocity of the ball = 0 (at maximum height, velocity is zero (0));

u = initial velocity of the ball = 42.4m/s

a = acceleration due to gravity = g = -10m/s² ( this is negative since the ball is thrown upwards against the direction of gravity)

t = time taken to reach maximum height = $t_{H}$

Substitute these values into equation (i) as follows;

0 = 42.4 - 10($t_{H}$)

$t_{H}$ = 42.4 / 10

$t_{H}$ = 4.24s

The time taken ($t_{H}$)to reach maximum height = 4.24s

Calculate the time taken to strike the ground from maximum height, $t_{G}$

(i) First let's get the maximum height reached relative to the roof of the building using another equation of motion;

h = ut + $\frac{1}{2}$ x a$t^{2}$      --------------------------(ii)

Where;

h = maximum height;

u = initial velocity of the ball = 42.4m/s

a = -10m/s² ( this is negative since the ball is thrown upwards against the direction of gravity)

t = time taken to reach maximum height = 4.24s

Substitute these values into equation (ii) as follows;

=> h = (42.4 x 4.24) - $\frac{1}{2}$ x 10 x 4.24²

=> h = 179.776 - 89.888

=> h = 89.888m

(ii) Second, let's find the time taken to strike the ground from maximum height using the same equation (ii);

Where;

h = total height relative to the ground = maximum height + building height

h = 89.888 + 68.1 = 157.988m

u = initial velocity from maximum height = 0 (at maximum height, velocity is 0)

a = acceleration due to gravity = +10m/s² (this is positive since the ball now moves downwards in the direction of gravity)

t = time taken from maximum height to strike the ground = $t_{G}$

Substitute these values into equation(ii);

157.988 = 0($t_{G}$) + $\frac{1}{2}$ x 10 x $t_{G}$²

157.988 = 5$t_{G}$²

Solve for $t_{G}$;

$t_{G}$² = 157.988 / 5

$t_{G}$² = 31.60

$t_{G}$ = $\sqrt{31.60}$

$t_{G}$ = 5.62s

Therefore, time taken to strike the ground from maximum height is 5.62s

Calculate the total time in air, $t_{T}$

$t_{T}$ = $t_{H}$ + $t_{G}$

$t_{T}$ = 4.24 + 5.62

$t_{T}$ = 9.86s

The total time the ball is in flight is 9.86s

Answer 2

Answer:

7.45 s.

Explanation:

Given:

h = 68.1 m

vi = 0 m/s

vf = 42.4 m/s

g = 9.81 m/s^2

Using,

h = vi*t +1/2*(a*t^2)

68.1 = 1/2 * (9.81*t^2)

t = sqrt((68.1*2)/9.81)

= 3.726 s.

Total time of flight = 2*t

= 2 * 3.726

= 7.45 s.