Question 92: Motion in 2D & Vectors

Question

The ranges and heights for two projectiles projected with the same initial velocity at angles 42$$^\circ$$ and 48$$^\circ$$ with the horizontal are R
1
, R
2
and H
1
, H
2
respectively. Choose the correct option :

Accepted Answer

Here, two projectiles are projected at angles 42$$^\circ$$ and 48$$^\circ$$ with same initial velocity.
As we know the expression of range of projectile,
Range $$ = {{{u^2}\sin 2	heta } \over g}$$
AT $$	heta$$
1
= 42$$^\circ$$,
Range, $${R_1} = {{{u^2}\sin 2(42)^\circ } \over g} = {{0.99{u^2}} \over g}$$
At $$	heta$$
2
= 48$$^\circ$$
Range, $${R_2} = {{{u^2}\sin 2(48)^\circ } \over g} = {{0.99{u^2}} \over g}$$
The range of the projectile is same for the two projectiles.
Therefore, R
1
= R
2
Now, as we know the expression of height of the projectile,
$${H_{\max }} = {{{u^2}\sin 	heta } \over {2g}}$$
At 42$$^\circ$$, $${H_{\max }} = {H_1} = {{{u^2}\sin 42^\circ } \over {2g}} = {{0.669{u^2}} \over {2g}}$$
At 48$$^\circ$$, $${H_{\max }} = {H_2} = {{{u^2}\sin 48^\circ } \over {2g}} = {{0.743{u^2}} \over {2g}}$$
Higher the value of $$	heta$$ higher the value of maximum height. Therefore, H
1
< H
2
.

Answer

R
1
> R
2
and H
1
= H
2

Answer

R
1
= R
2
and H
1
< H
2

Answer

R
1
< R
2
and H
1
< H
2

Answer

R
1
= R
2
and H
1
= H
2

Motion in 2D & Vectors

The ranges and heights for two projectiles projected with the same initial velocity at angles 42$^\circ and 48^\circ$ with the horizontal are R 1 , R 2 and H 1 , H 2 respectively. Choose the correct option :

Choose the correct answer: