Q83 PP - University Physics with Modern Physics

Question

About how long does it take a seed launched at 90° at the highest possible initial speed to reach its maximum height? Ignore air resistance. (a) 0.23 s; (b) 0.47 s; (c) 1.0 s; (d) 2.3 s.

Step-by-Step Solution

Answer

(b); It takes 0.47 s for the seed to reach the maximum height.

Step 1: Kinematic equation in terms of acceleration, velocity, and initial velocity.

For constant acceleration, the final velocityv is related to the initial velocityu by the expression given below,

$\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$

Step 2: Determine the time taken to reach the highest point.

The motion is vertical hence the kinematic equation is taken in terms of y-coordinates. As the motion is upwards the sign of g is taken to be negative.

$v_y=u_y+\left(-g\right)t$

The final velocity will be zero and the maximum initial velocity is 4.6 m/s, therefore the time taken to reach the highest point is

$$\begin{gathered} 0 = 4.6\frac{m}{s}- \left(9.8\frac{m}{s^2}\right)t \\ t = \frac{4.6{\displaystyle \frac{m}{s}}}{9.8{\displaystyle \frac{m}{s^2}}} \\ = 0.47 s \end{gathered}$$