Answer:
a) 29.36 m
b) 2.44 s
c) 2.57 s
d) 25.117 m/s
Step-by-step 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)
![v=u+at\\\Rightarrow 0=24-9.81* t\\\Rightarrow (-24)/(-9.81)=t\\\Rightarrow t=2.44 \s](https://img.qammunity.org/2020/formulas/physics/college/lj1fdrmf8mc1y6qirvle4qq5vl8rl6vugm.png)
Time taken by the ball to reach the highest point is 2.44 seconds
a)
![s=ut+(1)/(2)at^2\\\Rightarrow s=24* 2.44+(1)/(2)* -9.81* 2.44^2\\\Rightarrow s=29.35\ m](https://img.qammunity.org/2020/formulas/physics/college/1w8lhpyka91tslsndxp08jbsg1gzo5ys2l.png)
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
![s=ut+(1)/(2)at^2\\\Rightarrow 32.35=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(32.35* 2)/(9.81)}\\\Rightarrow t=2.57\ s](https://img.qammunity.org/2020/formulas/physics/college/12x4rj373i5u4851u2s3to8eevadsshanl.png)
The ball reaches the ground 2.57 seconds after reaching the highest point
d)
![v=u+at\\\Rightarrow v=0+9.81* 2.57\\\Rightarrow v=25.2117\ m/s](https://img.qammunity.org/2020/formulas/physics/college/57fzsp0d9cymyfgqq7lqvjcuxcegk0hlvs.png)
The ball will hit the ground at 25.2117 m/s