ANSWER
• The time it takes to land: ,0.63 seconds
,
• Horizontal distance: ,0.78 m
EXPLANATION
Let's draw a diagram of this situation,
The height of the seed is given by,
[tex]y=y_o+u_yt-\frac{1}{2}gt^2[/tex]
We want to know for what t, y = 0,
[tex]0=y_o+u_yt-\frac{1}{2}gt^2[/tex]
Before solving this equation, we have to find the vertical component of the initial velocity, u,
The vertical component is the opposite side to the angle the initial velocity forms with the horizontal, so we can find it using the sine of the angle,
[tex]\begin{gathered} \sin 62\degree=\frac{u_y}{u} \\ \\ u_y=u\cdot\sin 62\degree=2.65m/s\cdot\sin 62\degree\approx2.3398m/s \end{gathered}[/tex]
The information we have so far is:
• The vertical component of the initial velocity, uy = 2.3398 m/s
,
• The acceleration due to gravity, g = 9.81 m/s²
,
• The initial height of the seed, y₀ = 0.460 m
Replace in the equation for the height,
[tex]\begin{gathered} 0=0.46+2.3398t-\frac{1}{2}\cdot9.81\cdot t^2 \\ \\ 0=0.46+2.3398t-4.905\cdot t^2 \end{gathered}[/tex]
To solve this, we can use the quadratic formula,
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]
In this case, x = t, a = -4.905, b = 2.3398, and c = 0.46,
[tex]t=\frac{-2.3398\pm\sqrt[]{2.3398^2-4\cdot(-4.905)\cdot0.46}}{2\cdot(-4.905)}\approx\frac{-2.3398\pm\sqrt[]{14.49986}}{-9.81}\approx\frac{-2.3398\pm3.8079}{-9.81}[/tex]
There are two results,
[tex]\begin{gathered} t_1=\frac{-2.3398+3.8079}{-9.81}=\frac{1.4681}{-9.81}\approx-0.15 \\ \\ t_2=\frac{-2.3398-3.8079}{-9.81}=\frac{-6.1477}{-9.81}\approx0.63 \end{gathered}[/tex]
Remember that t represents time, so it cannot be negative.
Hence, the seed will land after 0.63 seconds.
The horizontal distance of a projectile is given by,
[tex]x=u_xt[/tex]
In projectile motion, the horizontal acceleration is 0, so the velocity is constant and it is the initial horizontal velocity.
Using the same reasoning we did to find the vertical component of the initial velocity, we now find the horizontal component (the adjacent leg of the angle) using the cosine of the angle,
[tex]\begin{gathered} \cos 62\degree=\frac{u_x}{u_{}} \\ \\ u_x=u\cos 62\degree=2.65m/s\cdot\cos 62\degree\approx1.2441m/s \end{gathered}[/tex]
The horizontal distance the seed covers in the 0.63 seconds we found before is,
[tex]x=1.2441m/s\cdot0.63s\approx0.78m[/tex]
Hence, the seed covers 0.78 meters horizontally before landing.
