Respuesta :
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.
Answer:
9.86s
Explanation:
The total time ([tex]t_{T}[/tex]) the ball is in flight is the sum of the time taken ([tex]t_{H}[/tex]) to reach maximum height and the time taken ([tex]t_{G}[/tex]) to strike the ground from maximum height. i.e
[tex]t_{T}[/tex] = [tex]t_{H}[/tex] + [tex]t_{G}[/tex]
Calculate time taken to reach maximum height, [tex]t_{H}[/tex].
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 = [tex]t_{H}[/tex]
Substitute these values into equation (i) as follows;
0 = 42.4 - 10([tex]t_{H}[/tex])
[tex]t_{H}[/tex] = 42.4 / 10
[tex]t_{H}[/tex] = 4.24s
The time taken ([tex]t_{H}[/tex])to reach maximum height = 4.24s
Calculate the time taken to strike the ground from maximum height, [tex]t_{G}[/tex]
(i) First let's get the maximum height reached relative to the roof of the building using another equation of motion;
h = ut + [tex]\frac{1}{2}[/tex] x a[tex]t^{2}[/tex] --------------------------(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) - [tex]\frac{1}{2}[/tex] 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 = [tex]t_{G}[/tex]
Substitute these values into equation(ii);
157.988 = 0([tex]t_{G}[/tex]) + [tex]\frac{1}{2}[/tex] x 10 x [tex]t_{G}[/tex]²
157.988 = 5[tex]t_{G}[/tex]²
Solve for [tex]t_{G}[/tex];
[tex]t_{G}[/tex]² = 157.988 / 5
[tex]t_{G}[/tex]² = 31.60
[tex]t_{G}[/tex] = [tex]\sqrt{31.60}[/tex]
[tex]t_{G}[/tex] = 5.62s
Therefore, time taken to strike the ground from maximum height is 5.62s
Calculate the total time in air, [tex]t_{T}[/tex]
[tex]t_{T}[/tex] = [tex]t_{H}[/tex] + [tex]t_{G}[/tex]
[tex]t_{T}[/tex] = 4.24 + 5.62
[tex]t_{T}[/tex] = 9.86s
The total time the ball is in flight is 9.86s
