Q84 PP - University Physics with Modern Physics

Question

If a seed is launched at an angle of 0° with the maximum initial speed, how far from the plant will it land? Ignore air resistance, and assume that the ground is flat. (a) 20 cm; (b) 93 cm; (c) 2.2 m; (d) 4.6 m.

Step-by-Step Solution

Answer

(b); The seed will land 0.92 m from the plant.

Step 1: Kinematic equations for projectile motion

For horizontal motion,

$$\begin{gathered} v_x=u_x \\ x=x_o+u_{x}t \end{gathered}$$

For vertical motion,

$$\begin{gathered} v_y=u_y+gt \\ y=y_o+u_{y}t+\frac{1}{2}g{t}^2 \end{gathered}$$

Step 2: Calculate the maximum horizontal distance covered by the seed.

Here the initial values of x and y value are zero. The maximum initial horizontal speed is 4.6 m/s. The initial speed in the vertical component is zero. The vertical height is 20 cm.Inserting this data into the vertical displacement equation

$$\begin{gathered} y=0+0+\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2y}{g}} \\ =\sqrt{\frac{2(2.0m)}{9.8 m/s^2}} \\ =0.20 s \end{gathered}$$

Insert this value in the horizontal displacement equation, and we get,

$$\begin{gathered} x = 0 + \left(4.6 m/s\right)\left(0.20 s\right) \\ = 0.92 m \end{gathered}$$

Hence, the seed will land 0.92 m from the plant.