Answer:
a) 29.36 m
b) 2.44 s
c) 2.57 s
d) 25.117 m/s
Explanation:
t = Time taken
u = Initial velocity = 24 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
b)
[tex]v=u+at\\\Rightarrow 0=24-9.81\times t\\\Rightarrow \frac{-24}{-9.81}=t\\\Rightarrow t=2.44 \s[/tex]
Time taken by the ball to reach the highest point is 2.44 seconds
a)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times 2.44+\frac{1}{2}\times -9.81\times 2.44^2\\\Rightarrow s=29.35\ m[/tex]
The highest point reached by the ball above its release point is 29.36 m
c) Total height is 3+29.35 = 32.35 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 32.35=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.35\times 2}{9.81}}\\\Rightarrow t=2.57\ s[/tex]
The ball reaches the ground 2.57 seconds after reaching the highest point
d)
[tex]v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2117\ m/s[/tex]
The ball will hit the ground at 25.2117 m/s
