13. A ball is thrown vertically upwards and it returns to the thrower after 6 sec ( g=9.8 m/s2) Find.
(i) The velocity with which it was thrown up. (ii) The maximum height it reaches
(iii) Its position after 4 sec.
To find the answers to these questions, we'll need to use the equations of motion for vertical motion. The three equations we'll use are:
1. v = u + gt
2. s = ut + (1/2)gt^2
3. v^2 = u^2 + 2gs
Where:
- v represents the final velocity
- u represents the initial velocity
- g is the acceleration due to gravity (9.8 m/s^2)
- t represents time
- s represents displacement (in this case, height)
(i) To find the initial velocity with which the ball was thrown up, we can use the first equation. In this case, the final velocity is 0 at the highest point of the trajectory. Thus, we have:
0 = u - (9.8 m/s^2)(6 s)
Rearranging the equation, we can solve for u:
u = (9.8 m/s^2)(6 s)
Therefore, the initial velocity with which the ball was thrown up is 58.8 m/s.
(ii) To find the maximum height the ball reaches, we can use the second equation. Again, the final velocity is 0 at the highest point, so:
0 = (58.8 m/s) * t - (1/2)(9.8 m/s^2)(t^2)
We can rearrange the equation to solve for t:
(1/2)(9.8 m/s^2)(t^2) = (58.8 m/s) * t
Simplifying further:
4.9 m/s^2 * t^2 - 58.8 m/s * t = 0
We can factor out t to solve for it:
t(4.9 m/s * t - 58.8 m/s) = 0
This equation gives us two potential solutions: t = 0 (when the ball is initially thrown) or 4.9 m/s * t - 58.8 m/s = 0. Since the ball returns to the thrower after 6 seconds, we can discard the t = 0 solution. Using the second solution, we get:
4.9 m/s * t - 58.8 m/s = 0
4.9 m/s * t = 58.8 m/s
t = 58.8 m/s / 4.9 m/s
t = 12 seconds
Substituting this value of t back into the second equation, we can find the maximum height:
s = (58.8 m/s)(12 s) - (1/2)(9.8 m/s^2)(12 s)^2
Simplifying:
s = 352.8 m - (1/2)(9.8 m/s^2)(144 s^2)
s = 352.8 m - 70.2 m/s^2 * s^2
s = 352.8 m - 10.098 s^2
Therefore, the maximum height the ball reaches is 352.8 m.
(iii) To find the position of the ball after 4 seconds, we can use the second equation:
s = (58.8 m/s)(4 s) - (1/2)(9.8 m/s^2)(4 s)^2
Simplifying:
s = 235.2 m - (1/2)(9.8 m/s^2)(16 s^2)
s = 235.2 m - 78.4 s^2
Therefore, the position of the ball after 4 seconds is 235.2 m.
13. Tr = 6/2 = 3 s. = Rise time.
Tf = Tr = 3 s. = Fall time.
a. V = Vo + g*Tr = 0 @ max h.
Vo = -g*Tr = -(-9.8)*3 = 29.4 m/s. =
Initial velocity.
b. h = Vo*Tr + 0.5g*Tr^2.
h = 29.4*3 - 4.9*3^2 = 44.1 m.
c. Since Tr is 3 seconds, the ball is
on its' way down after j seconds: Tf = 4-3 = 1 s.
h = ho-0.5g*Tf^2 = 44.1 - 4.9*1^2 = 39.2 m. Above launching point.