Formative Practice 2.3 (Question 3 & 4) – Physics Form 4 Chapter 2 (Force and Motion I)

Question 3:
An object thrown vertically upwards reached a maximum height of 5.0 m. Calculate:
(a) the velocity of the object when thrown
(b) the time taken for the object to reach its maximum height
(c) the time required for the object to return to its original level
Ignore air resistance. [g = 9.81 m s^–2]

Answer:
(a)

$$\begin{aligned} & v=0 \mathrm{~ms}^{-1} \\ & s=+5 \mathrm{~m} \\ & g=-9.81 \mathrm{~ms}^{-2} \\ & u=? \end{aligned}$$
$$\begin{aligned} v^2 & =u^2+2 g s \\ 0^2 & =u^2+2(-9.81)(5) \\ u^2 & =98.1 \\ u & =9.90 \mathrm{~ms}^{-1} \end{aligned}$$

(b)

$$\begin{aligned} u & =9.9 \mathrm{~ms}^{-1} \\ v & =0 \mathrm{~ms}^{-1} \\ g & =-9.81 \mathrm{~ms}^{-2} \\ t & =? \end{aligned}$$
$$\begin{aligned} v & =u+g t \\ 0 & =9.9+(-9.81) t \\ 9.81 t & =9.9 \\ t & =1.01 \mathrm{~s} \end{aligned}$$

(c)
The time to descend to the original position is the same as the time to ascend.
$$\begin{aligned} \therefore t_1+t_2 & =1.01+1.01 \\ & =2.02 \mathrm{~s} \end{aligned}$$

Question 4:
A tennis ball that is released falls vertically from a building of height 50 m. Calculate:
(a) the time taken for the ball to reach the base of the building
(b) the velocity of the ball just before hitting the base of the building
(c) the vertical distance passed at the third second
Ignore air resistance. [g = 9.81 m s^–2]

Answer:
$$\begin{aligned} u & =0 \mathrm{~ms}^{-1} \\ s & =50 \mathrm{~m} \\ a & =9.81 \mathrm{~ms}^{-2} \text { (assuming downward } \\ t & =? \quad \text { movement is positive) } \end{aligned}$$

(a)
$$\begin{aligned} s & =u t+\frac{1}{2} g t^2 \\ 50 & =0+\frac{1}{2}(9.81) t^2 \\ t^2 & =\frac{50 \times 2}{9.81} \\ t & =3.19 \mathrm{~s} \end{aligned}$$

(b)
$$\begin{aligned} v & =u+g t \\ & =(0)^2+(9.81)(3.19) \\ & =31.29 \mathrm{~ms}^{-1} \end{aligned}$$

(c)
$$\begin{aligned} &\text { The } 3 \text { rd second }=t_3-t_2\\ &\begin{aligned} s & =\left(u t+\frac{1}{2} g t_3{ }^2\right)-\left(u t+\frac{1}{2} g t_2^2\right) \\ s & =\left[0+\frac{1}{2}(9.81)(3)^2\right]-\left[0+\frac{1}{2}(9.81)(2)^2\right] \\ & =44.15-19.62 \\ & =24.53 \mathrm{~m} \end{aligned} \end{aligned}$$