AP EAMCET · PHYSICS · Motion In Two Dimensions
A body is projected from the top of a tower with a velocity \(\overrightarrow{\mathbf{u}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \mathrm{ms}^{-1}\), where \(\hat{\mathbf{i}}, \hat{\mathbf{j}}\) and \(\hat{\mathbf{k}}\) are unit vectors along east, north and vertically upwards respectively. If the height of the tower is \(30 \mathrm{~m}\), horizontal range of the body on the ground is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
- A 15 m
- B 25 m
- C 9 m
- D 12 m
Answer & Solution
Correct Answer
(A) 15 m
Step-by-step Solution
Detailed explanation
( \(\hat{\mathbf{k}}\) is given vertically upward direction) \( \begin{aligned} & a=-10 \mathrm{~m} / \mathrm{s}^2 \\ & h=-30 \mathrm{~m} \end{aligned} \) As, \(S=u t+\frac{1}{2} a t^2 \Rightarrow-30=5 t-\frac{1}{2} \times 10 \times t^2\) \( \Rightarrow t^2-t-6=0 \)…
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