In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is 20 cm tall, and the beetle jumps with an initial velocity of 1.5m/s at an angle of 20∘ above the horizontal. How far from the base of the step does the beetle land on the ground?

Respuesta :

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

[tex]\theta[/tex] = Angle = 20°

y = -20 cm

Velocity components

[tex]u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s[/tex]

[tex]u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s[/tex]

Acceleration components

[tex]a_x=0[/tex]

[tex]a_y=-9.81\ m/s^2[/tex]

[tex]y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0[/tex]

[tex]t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629[/tex]

Time taken is 0.26088 seconds

[tex]x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m[/tex]

The distance the beetle travels on the ground is 0.3677181864 m