Kinematics Question 632
Question: You throw a ball with a $ \vec{v}=( 3\hat{i}+4\hat{j} )\text{m/s} $ towards a wall, where it his at height $ h _{1} $ . Suppose that the launch velocity were, instead, $ \vec{v}=( 5\hat{i}+4\hat{j} )\text{m/s} $ and $ h _{2} $ it’s height, then
Options:
A) $ {{\text{h}} _{\text{1}}}\text{=}{{\text{h}} _{\text{2}}} $
B) $ {{\text{h}} _{\text{2}}}\text{}{{\text{h}} _{\text{1}}} $
C) $ {{\text{h}} _{\text{2}}}\text{}{{h _{1}} $
D)$ {{\text{h}} _{\text{2}}}\ge {{\text{h}} _{\text{1}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ x=3t _{1}=5t _{2}\text{}\Rightarrow t _{1}=x/3\text{ and }t _{2}=x/5 $
Now $ h _{1}=4t _{1}-\frac{1}{2}gt _{1}^{2}=\frac{4}{3}x-\frac{gx^{2}}{18}. $
and $ h _{2}=4t _{2}-\frac{1}{2}gt _{2}^{2}=\frac{4x}{5}-\frac{gx^{2}}{10}. $
Clearly, $ h _{2}<h _{1}. $
