Respuesta :
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
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[tex]m/s^{2}[/tex], 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 + [tex]\frac{1}{2}[/tex]a[tex]t^{2}[/tex]
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 - [tex]\frac{1}{2}[/tex] x g x [tex]1.6^{2}[/tex]
Taking g as 10[tex]m/s^{2}[/tex], we have;
h = 16x1.6 - [tex]\frac{1}{2}[/tex] x 10 x [tex]1.6^{2}[/tex]
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 + [tex]\frac{1}{2}[/tex]a[tex]t^{2}[/tex]
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 + [tex]\frac{1}{2}[/tex] x g x [tex]t^{2}[/tex]
Taking g as 10[tex]m/s^{2}[/tex], we have;
12.8025 = 16xt + [tex]\frac{1}{2}[/tex] x 10 x [tex]t^{2}[/tex]
12.8025 = 16t + 5[tex]t^{2}[/tex]
Rearranging;
5[tex]t^{2}[/tex] + 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.
