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NSEP 2019 Problem 26 | Projectile Motion

Try this beautiful 2D Motion Problem based on Projectile Motion from NSEP - 2019.

2D Motion Problem - NSEP 2019 Problem 26


A particle is projected from the ground with a velocity $\vec{V}$=$(3 \hat{i}+10 \hat{j}) \mathrm{ms}^{-1}$. The maximum height attained and the range of the particle are respectively given by (use $\mathrm{g}$=$10 \mathrm{~m} / \mathrm{s}^{2}$ )
(a) $5 \mathrm{~m}$ and $6 \mathrm{~m}$
(b) $3 \mathrm{~m}$ and $10 \mathrm{~m}$
(c) $6 \mathrm{~m}$ and $5 \mathrm{~m}$
(d) $3 \mathrm{~m}$ and $5 \mathrm{~m}$

Key Concepts


Kinematics

2 Dimensional Motion

Projectile Motion

Suggested Book | Source | Answer


Resnick Halliday Krane

NSEP 2019; Question no:26

(a) $5 \mathrm{~m}$ and $6 \mathrm{~m}$

Try with Hints


Resolution of the Components of Vectors

For the Vertical journey which component of the Velocity is responsible for it

${H}=\frac{\mathrm{u}_{\mathrm{y}}^{2}}{2 \mathrm{~g}}=\frac{10^{2}}{20}=5 \mathrm{~m}$

Could we able to think about the relation between the Range and the horizontal component of Velocity

$R=u_{x} \frac{2 u_{y}}{g}=3 \times \frac{2 \times 10}{10}=6 \mathrm{~m}$

Physics Olympiad Program at Cheenta

Subscribe to Cheenta at Youtube


Try this beautiful 2D Motion Problem based on Projectile Motion from NSEP - 2019.

2D Motion Problem - NSEP 2019 Problem 26


A particle is projected from the ground with a velocity $\vec{V}$=$(3 \hat{i}+10 \hat{j}) \mathrm{ms}^{-1}$. The maximum height attained and the range of the particle are respectively given by (use $\mathrm{g}$=$10 \mathrm{~m} / \mathrm{s}^{2}$ )
(a) $5 \mathrm{~m}$ and $6 \mathrm{~m}$
(b) $3 \mathrm{~m}$ and $10 \mathrm{~m}$
(c) $6 \mathrm{~m}$ and $5 \mathrm{~m}$
(d) $3 \mathrm{~m}$ and $5 \mathrm{~m}$

Key Concepts


Kinematics

2 Dimensional Motion

Projectile Motion

Suggested Book | Source | Answer


Resnick Halliday Krane

NSEP 2019; Question no:26

(a) $5 \mathrm{~m}$ and $6 \mathrm{~m}$

Try with Hints


Resolution of the Components of Vectors

For the Vertical journey which component of the Velocity is responsible for it

${H}=\frac{\mathrm{u}_{\mathrm{y}}^{2}}{2 \mathrm{~g}}=\frac{10^{2}}{20}=5 \mathrm{~m}$

Could we able to think about the relation between the Range and the horizontal component of Velocity

$R=u_{x} \frac{2 u_{y}}{g}=3 \times \frac{2 \times 10}{10}=6 \mathrm{~m}$

Physics Olympiad Program at Cheenta

Subscribe to Cheenta at Youtube


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