An athlete throws a javelin at an angle of $45^{\circ}$ to the horizontal and, after a time interval $t$, the javelin hits the ground $16$ $\text{m}$ in front of the throwing point . Neglecting the air resistance and the athlete's height, the time interval $t$, in seconds, is a value closer to? (given $\sin 45^{\circ} = \cos 45^{\circ} = 0.7$)
Attempt: We have that $v_x=v_y=0.7 v$. For reach (assuming that the throw has symmetrical height and reach at the same level) the reach it took the ball to reach the maximum height will be equal to the total reach, so $$8 = v_{0_x} t \implies t = \frac{0.8}{0.7 v_0}$$ Now, at the maximum height we have that $$0 = v_{0_y} - 10 t \implies t = \frac{40 \sqrt{5}}{7}$$ So the time will be $2t = \frac{80 \sqrt{5}}{7}$ The answer to this question is $16$. Where am I wrong?
You have (assuming $g=9.8\ m/s^2$):
$$d_x = v_{0x}t$$ $$d_y = v_{0y}t-4.9t^2$$
Setting $d_y$ to zero (where the javelin is at the horizontal axis) gives $t=0,\dfrac{v_{0y}}{4.9}$. Clearly it is the positive value. Then
$$16=d_x = v_{0x}t=v_{0x}\cdot \dfrac{v_{0y}}{4.9} = \dfrac{\left(\dfrac{\sqrt{2}}{2}v_0\right)^2}{4.9}=\dfrac{v_0^2}{9.8}$$
So $$v_0 = \sqrt{16\cdot 9.8}$$ $$t = \dfrac{d_x}{v_{0x}} = \dfrac{16}{{\sqrt{2}\over {2}} \cdot \sqrt{16\cdot 9.8}}\approx 1.807$$
You will arrive at a similar result using $0.7$ instead of $\sqrt 2\over 2$.