Respuesta :
Answer:
a) 2.55 seconds
b) 31.86 m
c) 1.4 seconds
Explanation:
t = Time taken
u = Initial velocity = 25 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
a)
[tex]v=u+at\\\Rightarrow 0=25-9.81\times t\\\Rightarrow \frac{-25}{-9.81}=t\\\Rightarrow t=2.55 \s[/tex]
Time taken by the ball to reach the highest point is 2.55 seconds
b)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=25\times 2.55+\frac{1}{2}\times -9.81\times 2.55^2\\\Rightarrow s=31.86\ m[/tex]
The highest point reached by the ball above its release point is 31.86 m
c)
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 25+25^2}\\\Rightarrow v=11.6\ m/s[/tex]
[tex]v=u+at\\\Rightarrow 11.6=25-9.81\times t\\\Rightarrow \frac{11.6-25}{-9.81}=t\\\Rightarrow t=1.4 \s[/tex]
It takes 1.4 seconds to reach a point of 25 m above its release point
