Q45P - Fundamentals Of Physics

Question

(a) with what speed must a ball be thrown vertically from ground level to rise to a maximum height of? (b)How long will it be in the air? (c)Sketch graphs of y, v, a vs. t for the ball. On the first two graphs, indicate the time at which 50 m is reached.

Step-by-Step Solution

Answer

1. Initial speed for the ball to travel 50 m vertically is 31.3m/s
2. Time duration for the ball in the air 6.4 s

Step 1: Given information

The given information is,

$$\begin{gathered} \mathrm{Acceleration} \left(\mathrm{a}\right)=\mathrm{Gravitational} \mathrm{Acceleration}=9.8 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Displacement} \left(\mathrm{y}\right)=50 \mathrm{m} \\ \mathrm{Initial} \mathrm{Velocity} \left({\mathrm{v}}_0\right)=0 \mathrm{m}/\mathrm{s} \end{gathered}$$

Step 2: Understanding the concept

The problem deals with the concept of free fall. In free fall the object experiences the motion where gravity is the only force acting upon it. This concept can be used to find the velocity at which the ball must be thrown to reach a certain height. Once thrown, it would be acted upon by gravitational acceleration only. Hence, at maximum height,, the final velocity would be zero. Using these conditions, the initial velocity and the time for which ball would be in the air can be found.

Formula:

$$\begin{gathered} v^2=v_0^2+2ay \\ v=v_0+at \end{gathered}$$

Step 3: (a) Determination of initial speed of ball

Using the third equation for kinematic, 

$$\begin{gathered} v^2={v_0}^2+2 ay \\ {v}^2=0+2\times 9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\times 50 \mathrm{m} \\ {v}^2=\sqrt{980} {\mathrm{m}}^2/{\mathrm{s}}^2 \\ v=31.3 \mathrm{m}/\mathrm{s} \end{gathered}$$

Initial speed for the ball to travel 50mvertically is 31.3 m/s

Step 4: (b) Determination of time taken by ball in the air

The total time taken by the ball to reach the maximum height can be found as below

$v=v_0+at$

 The final velocity at the maximum height would be zero. “a” is the acceleration, and in this case, it is taken as “- g”. It is negative because the upward direction is taken as positive.

$$\begin{gathered} 0=v_0-gt \\ t=v/g \end{gathered}$$

But the time duration of the ball in the air, so if it takes ‘t’ seconds to reach at maximum height, it will take ‘t’ seconds to reach to ground from the maximum height.

So

$$\begin{gathered} T=t+t \\ =2t \\ T=2\times \frac{v}{g} \\ T=2\times \frac{31.3 \mathrm{m}/\mathrm{s}}{9.8 \mathrm{m}/{\mathrm{s}}^2} \\ T=6.38 \\ \approx 6.4 \sec \end{gathered}$$

Step 5: Graph between displacement vs time

Step 6: Graph between velocity vs time

Step 7: Graph between acceleration vs time