Answer:
Time taken by the arrow to travel along to hit the ground is 0.55 seconds.
Explanation:
The only "force" acting on the "crossbow" to cause it to "hit" the ground is "gravity". There is no initial velocity downward when it shoot.
[tex]d=v_{i} t+\frac{1}{2} t^{2}[/tex]
d = the displacement of the object
t = the time for which the object moved
a = acceleration of the object
[tex]v_i[/tex] = the initial velocity of the object
Given values
d = 1.5 m
t = unknown
[tex]a=g=9.8 \mathrm{m} / \mathrm{s}^{2}[/tex]
[tex]\mathrm{V}_{\mathrm{i}}=0 \mathrm{m} / \mathrm{s}[/tex]
[tex]1.5 \mathrm{m}=0(\mathrm{t})+\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) \mathrm{t}^{2}[/tex]
[tex]1.5=0+4.9 \mathrm{t}^{2}[/tex]
[tex]\mathrm{t}^{2}=\frac{1.5}{4.9}[/tex]
[tex]t^{2}=0.306 \mathrm{s}[/tex]
Square root both sides
[tex]t=\sqrt{0.306}[/tex]
t = 0.55 s
