Question

1. An airow is shot horizontally from a crossbow 1.5 m above the ground. The initial velocity is 45

Answer 1

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.

$d=v_{i} t+\frac{1}{2} t^{2}$

d = the displacement of the object  

t = the time for which the object moved  

a = acceleration of the object  

$v_i$ = the initial velocity of the object

Given values

d = 1.5 m

t = unknown

$a=g=9.8 \mathrm{m} / \mathrm{s}^{2}$

$\mathrm{V}_{\mathrm{i}}=0 \mathrm{m} / \mathrm{s}$

$1.5 \mathrm{m}=0(\mathrm{t})+\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) \mathrm{t}^{2}$

$1.5=0+4.9 \mathrm{t}^{2}$

$\mathrm{t}^{2}=\frac{1.5}{4.9}$

$t^{2}=0.306 \mathrm{s}$

Square root both sides

$t=\sqrt{0.306}$

t = 0.55 s