Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The distances and heights of two projectiles fired at the same initial velocity at angles $42$ and to $48$ the

horizontal is respectively $R_1,R_2$ and $H_1,\ H_2$ .
Option: 1
$R_1>R_2\ and\ H_1=H_2$

Option: 2
$R_1=R_2\ and\ H_1<H_2$

Option: 3
$R_1<R_2\ and\ H_1<H_2$

Option: 4
$R_1=R_2\ and\ H_1=H_2$

Answers (1)

best_answer

Answer: (B) $R_1=R_2\ and\ H_1<H_2$

Explanation:

Here are two projectiles launched at angles $42^{\circ}$ and $48^{\circ}$ with the same initial speed.
As we know the expression for projectile range,
$Range\ =\frac{u^2 sin 2\theta}{g}$
At

$\theta_1=42^{\circ}$

Range, $R_1=\frac{u^2 \sin2(42^{\circ})}{g}$

At,
$\theta_2=48^{\circ}$
$R_2=\frac{u^2sin2\left (48^{\circ} \right )}{g}$
Therefore, $R_1=R_2$

Now, we know the expression for the height of the projectile.

$H_{max}=\frac{u^2 sin\theta}{2g}$
For

$42^{\circ}$
$H_{max}=H_1=\frac{u^2sin\left (42^{\circ} \right )}{2g}=\frac{0.669u^2}{2g}$

At $48^{\circ}$ ,

$H_{max}=H_2=\frac{u^2sin 2\left (48^{\circ} \right )}{2g}=\frac{0.743u^2}{2g}$

Higher maximum height value. Therefore, $H1 < H2$ .

Posted by

shivangi.shekhar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks

anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule

anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question