Answer:
0.95 seconds
Explanation:
t = Time taken
u = Initial velocity = 15 m/s
v = Final velocity
s = Displacement
a = Acceleration = 9.81 m/s² (downward positive, upward negative)
Time taken by the ball to reach the maximum height
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-15}{-9.81}\\\Rightarrow t=1.52\ s[/tex]
Maximum height
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-15^2}{2\times -9.81}\\\Rightarrow s=11.47\ m[/tex]
Distance between maximum height of the ball and the branch is 11.47-7 = 4.46 m
So, the distance that will be covered on the way down is 4.46 m
Now
u = 0
s = 4.46
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 4.46=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4.46\times 2}{9.81}}\\\Rightarrow t=0.95\ s[/tex]
Time taken by the ball from the maximum height to the tree branch is 0.95 seconds.
Total time taken from the moment the ball is thrown to reach the tree branch is 1.52+0.95 = 2.47 seconds
