Answer:
Explanation:
Given
Initial velocity of ball [tex]u=10\ m/s[/tex]
height of window [tex]h=20\ m[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where u=initial velocity
t=time
a=acceleration
As ball is already is at a height of 20 m so
[tex]Y=ut+\frac{1}{2}at^2+20[/tex]
[tex]Y=10\times t+0.5\times (-9.8)t^2+20[/tex]
[tex]Y=-4.9t^2+10t+20[/tex]
(b)highest point is obtained at v=0
[tex]v^2-u^2=2as[/tex]
where
v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](0)-10^2=2\times (-9.8)\times s[/tex]
[tex]s=\frac{100}{19.6}[/tex]
[tex]s=5.102\ m[/tex]
Highest Point will be [tex]s+20=25.102\ m[/tex]
(c)Time taken when the ball hit the ground i.e. at Y=0
[tex]-4.9t^2+10t+20=0[/tex]
[tex]t=3.28\ s[/tex]
impact velocity [tex]v=\sqrt{2\times 9.8\times 25.102}[/tex]
[tex]v=22.181\ m/s[/tex]
(a) The equation be "Y = -4.9t² + 10t + 20".
(b) The highest point be "25.102 m".
(c) The impact velocity be "22.181 m/s"
According to the question,
Ball's initial velocity, u = 10 m/s
Window's height, h = 20 m
(a) By using equation of motion,
Y = ut + [tex]\frac{1}{2}[/tex]at²
By substituting the values,
= ut + [tex]\frac{1}{2}[/tex]at² + 20
= 10 × t + 0.5 × (9.8)t² + 20
= -4.9t² + 10t + 20
(b) We know that,
→ v² - u² = 2as
here, Final velocity, v = 0
0 - (10)² = 2 × (-9.8) × s
s = [tex]\frac{100}{19.6}[/tex]
= 5.102 m
(c) Time taken will be:
→ -4.9t² + 10t + 20 = 0
t = 3.28 s
hence,
The impact velocity,
v = [tex]\sqrt{2\times 9.8\times 25.102}[/tex]
= 22.181 m/s
Thus the above response is correct.
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