Question 13 Marks
A stone is thrown vertically upwards with a speed of $20 \mathrm{~m} / \mathrm{s}$. How high will it go before it begins to fall? $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)$
Answer
View full question & answer→Initial velocity, $\mathrm{u}=20 \mathrm{~m} / \mathrm{s}$
Final velocity, $\mathrm{v}=0$
Acceleration due to gravity, $\mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^2 \mathrm{Height}, \mathrm{h}=$ ?
Using relation, for a freely falling body: $\mathrm{v}^2=\mathrm{u}^2+2 \mathrm{gh}(0)^2=(20)^2+2 \times(-9.8) \times \mathrm{h}\ 0-400=-19.6 \mathrm{~h} \mathrm{~h}=\frac{400}{19.6}=20.4 \mathrm{~m}$
Final velocity, $\mathrm{v}=0$
Acceleration due to gravity, $\mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^2 \mathrm{Height}, \mathrm{h}=$ ?
Using relation, for a freely falling body: $\mathrm{v}^2=\mathrm{u}^2+2 \mathrm{gh}(0)^2=(20)^2+2 \times(-9.8) \times \mathrm{h}\ 0-400=-19.6 \mathrm{~h} \mathrm{~h}=\frac{400}{19.6}=20.4 \mathrm{~m}$