Question

In case of a projectile motion, what is the angle
between the velocity and acceleration at the highest
point?
(a) 0°
(b) 45°
(c) 90°
(d) 180°​

Answer 1

Answer:

90°

Explanation:

As we know ,

In the projectile motion , at the topmost point ,

only horizontal velocity remains

And the acceleration due to gravity always acts

downwards

Both are mutually perpendicular

Answer 2

Solution:-

We know that To find maximum height using this formula

$\boxed{\rm \: H = \dfrac{ {u}^{2} \sin ^{2} \alpha }{2g} }$

Now take

Option:- A when  α = 0⁰

$\rm \: H = \dfrac{ {u}^{2} \sin ^{2} 0 {}^{o} }{2g}$

$\rm \to \: \sin0 {}^{o } = 0$

$\rm \: H = \dfrac{ {u}^{2} \times 0 {}^{} }{2g} = 0$

Option :- B when  α = 45⁰

$\rm \: H = \dfrac{ {u}^{2} \sin ^{2} 45 {}^{o} }{2g}$

$\rm \to \sin45 {}^{o} = \dfrac{1}{ \sqrt{2} }$

$\rm \: H = \dfrac{ {u}^{2} \times (\dfrac{1}{ \sqrt{2} }) ^{2} }{2g} = \dfrac{0.5 \times {u}^{2} }{2g}$

Option:- C when  α= 90⁰

$\rm \: H = \dfrac{ {u}^{2} \sin ^{2} 90 {}^{o} }{2g}$

$\rm \to \sin90 {}^{o} = 1$

$\rm \: H = \dfrac{ {u}^{2} \times 1{}^{} }{2g} = \dfrac{ {u}^{2} }{2g}$

Option:- D when  α = 180⁰

$\rm \: H = \dfrac{ {u}^{2} \sin ^{2} 180 {}^{o} }{2g}$

$\rm \to \sin180 {}^{o} = 0$

$\rm \: H = \dfrac{ {u}^{2} \times 0 {}^{} }{2g} = 0$

we clearly seen maximum height will be at 90⁰

So answer is 90⁰ option c is correct