Answer:
2.03873 s
70.38735 m
4.07747 seconds
5.82688 seconds
27.37482 m from the ground
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-20}{-9.81}\\\Rightarrow t=2.03873\ s[/tex]
time needed for the ball to reach its maximum height is 2.03873 s
The time taken to go up and the time taken to reach the point from where it was thrown is the same.
So, time needed for the ball to return to the height from which it was thrown is 2.03873+2.03873 = 4.07747 seconds
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -9.81}\\\Rightarrow s=20.38735\ m[/tex]
The maximum height the ball will reach is 50+20.38735 = 70.38735 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 70.38735=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{70.38735\times 2}{9.81}}\\\Rightarrow t=3.78815\ s[/tex]
Time needed to reach the ground is 2.03873+3.78815 = 5.82688 seconds
The time from the maximum height that is required is 5-2.03873 = 2.96127 seconds
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.96127^2\\\Rightarrow s=43.01253\ m[/tex]
The ball will be 70.38735-43.01253 = 27.37482 m from the ground
