KCET · Physics · Motion In Two Dimensions
The height \(y\) and the distance \(x\) along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by \(y=8 t-5 t^{2} \mathrm{~m}\) and \(x=6 t \mathrm{~m}\), where \(t\) is in seconds. The velocity with which the projectile is projected is
- A \(6 \mathrm{~ms}\)
- B \(8 \mathrm{~ms}^{1}\)
- C \(10 \mathrm{~ms}^{-1}\)
- D \(14 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(10 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, \(\quad y=8 t-5 t^{2} \quad \ldots\) (i)
We know,
\(x=6 t\)
Compare with Eq. (ii), we get
\(u_{1} \cos \theta=\frac{x}{t}=6\)
and
\(y=(u \sin \theta) t-\frac{1}{2} g t^{2}\)
Compare with Eq (i), we get
\(\begin{aligned}u_{2} \sin \theta &=8 \\u &=\sqrt{u_{1}^{2}+u_{2}^{2}} \\u &=\sqrt{36+64} \\u &=10 \mathrm{~ms}^{-1}\end{aligned}\)
We know,
\(x=6 t\)
Compare with Eq. (ii), we get
\(u_{1} \cos \theta=\frac{x}{t}=6\)
and
\(y=(u \sin \theta) t-\frac{1}{2} g t^{2}\)
Compare with Eq (i), we get
\(\begin{aligned}u_{2} \sin \theta &=8 \\u &=\sqrt{u_{1}^{2}+u_{2}^{2}} \\u &=\sqrt{36+64} \\u &=10 \mathrm{~ms}^{-1}\end{aligned}\)
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