Answer:
Explanation:
Given :
Distance in the vertical direction, S = 56 cm
= 0.56 m
Initial velocity in the vertical direction, u = 0 m/s
Acceleration due to gravity, g = 9.81 [tex]m/s^2[/tex]
Therefore, in order to find the time required,
[tex]$S = ut + \frac{1}{2}at^2$[/tex]
[tex]$0.56 = o(t) + \frac{1}{2}(9.81)t^2$[/tex]
[tex]$t^2=\frac{0.56 \times 2}{9.81}$[/tex]
[tex]$t^2=0.114$[/tex]
t = 0.33 seconds
Therefore, the initial velocity of the arrow is given by :
[tex]$v_0=\frac{S}{t}$[/tex] , where S = 11 m and t = 0.33 seconds
[tex]$v_0=\frac{11}{0.33}$[/tex]
[tex]$v_0=33.33\ m/s$[/tex]
Thus the initial velocity of the arrow is 33.33 m/s.
