Question

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0 m/sm/s when the hand is 2.50 mm above the ground. How long is the ball in the air before it hits the ground? ( The student moves her hand out of the way)

Answer 1

Answer:

Time in air = 2.26 seconds

Explanation:

Time in air is the sum of the time taken to reach maximum height and the time taken to hit the ground from maximum height.

i.e

Time in air = Time to reach max. height + Time to hit the ground from max. height

Calculate time to reach max height

Using one of the equations of motion;

v = u + at

where;

v = final velocity at max. height = 0 (at max. height, final velocity is zero)

u = initial velocity = 16.0m/s

a = acceleration due to gravity = -g (since the ball moves up against gravity)

t = time taken

Substitute the values of v, u and a into the equation above;

0 = 16 - gt

Taking g as 10$m/s^{2}$, we have;

0 = 16 -10t

Making t subject of the formula;

t = 16 / 10 = 1.6 s

Therefore, time taken to reach max. height is 1.6 seconds

Calculate time to hit ground from max. height

First calculate the max. height reached using the equation;

h = ut + $\frac{1}{2}$a$t^{2}$

where;

h is the height

u = initial velocity = 16m/s

a = acceleration due to gravity = -g (since the ball moves up against gravity)

t = time taken to reach max. height = 1.6s

Substitute the values of u, a and t into the equation above;

h = 16x1.6 - $\frac{1}{2}$ x g x $1.6^{2}$

Taking g as 10$m/s^{2}$, we have;

h = 16x1.6 - $\frac{1}{2}$ x 10 x $1.6^{2}$

h = 25.6 - 12.8

h = 12.8m

Remember that the ball was thrown at a height of 2.50mm above the ground;

Therefore, total height reached = 12.8m + 2.50mm

Convert 2.50mm to m we have

=> 2.50mm = 0.0025m

Total height reached = 12.8 + 0.0025 = 12.8025m

Now calculate the time taken to hit the ground using the same equation;

h = ut + $\frac{1}{2}$a$t^{2}$

where;

h is the total height reached = 12.8025m

u = initial velocity = 16m/s

a = acceleration due to gravity = +g (since the ball now moves downwards in the direction of gravity)

t = time taken to hit the ground

Substitute the values of u, a and t into the equation above;

12.8025 = 16xt + $\frac{1}{2}$ x g x $t^{2}$

Taking g as 10$m/s^{2}$, we have;

12.8025 = 16xt + $\frac{1}{2}$ x 10 x $t^{2}$

12.8025 = 16t + 5$t^{2}$

Rearranging;

5$t^{2}$ + 16t - 12.8205 = 0

Solving the quadratic equation gives;

t = 0.66 or t = -3.86

Since time cannot be negative, t = 0.66s

Therefore time taken to hit the ground is 0.66s

Calculate time in air

Time in air = Time to reach max. height + Time to hit the ground from max. height.

Time in air = 1.60 + 0.66

Time in air = 2.26seconds

Note: I'd imagine that the ball left the student's hand with when the hand was 2.50m above the ground and not 2.50mm. But then, i have followed what was typed. I have stuck to 2.50mm.

Answer 2

Answer:

The answer to the question is

The ball in the air for 3.51 s before it hits the ground

Explanation:

To solve the question. we list the variables

Intitial velocity = 16.0 m/s,

Initial height of ball from the ground = 2.5 m

acceleration due to gravity = 9.81 m/s²

The required relation is

v = u + at

where a = -g and g = acceleration due to gravity×

g = 9.81 m/s², v= 0 at top before coming back down

u = 16.5 m/s, thus 16.5 m/s = t × 9.8 m/s²

t = 16.5 ÷9.81 = 1.68 s

The ball height at top of its motion is =

S = ut - 0.5×g×t²

= 16.5×1.68 - 0.5×9.81×1.68²

= 13.876 m

Therefore total height 13.876 m+  2.5 m =  16.376 m

and the time it takes bacjk down is given by

S = ut + 0.5gt² ⇒ 16.376 m  = 0×t + 0.5×9.81×t²

16.376 m = t²×4.905 m/s² or t² = 3.339 s² or t = 1.827 s

Therefore total time = 1.807 s+ 1.68 s = 3.51 s

The ball in the air for  3.51 s