# if the kinetic energy of an oblique projectile

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## The potential energy of a projectile at maximum height is 34 times kinetic energy of projection. Its angle of projection is:

Click here👆to get an answer to your question ✍️ 9. If the kinetic energy of an oblique projectile at its maximum height is half of its initial kinetic energy then the angle of throw with the vertical is (2) 37 (3) 45- (4) 60\"

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## 9. If the kinetic energy of an oblique projectile at its maximum height is half of its initial kinetic energy then the angle of throw with the vertical is (2) 37 (3) 45- (4) 60"

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Updated on : 2022-09-05

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## If the kinetic energy of an oblique projectile at its maximum height is half of its initial kinetic energy then the angle of throw with the vertical is

If the kinetic energy of an oblique projectile at its maximum height is half of its initial kinetic energy then the angle of throw with the vertical is

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SIMILAR QUESTIONS

**Q.**The kinetic energy of a projectile at its highest position is

K

. If the range of the projectile is four times the height of the projectile, then the initial kinetic energy of the projectile is

**Q.**If we throw a body vertically upwards with velocity of 4 m/s. Then at what height does its kinetic energy reduce to half of initial value?(Take g=10 m/s-2)

**Q.**The kinetic energy of a projectile at the highest point is half of the initial kinetic energy. What is the angle of projection with the horizontal?

**Q.**The kinetic energy of a projectile at the highest point of its path is found to be

3 / 4 t h

of its initial kinetic energy. If the body is projected from the ground, the angle of projection must be:

**Q.**A projectile is projected with a kinetic energy

K

. If it has the maximum possible horizontal range, then its kinetic energy at the highest point will be

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## If the kinetic energy of a projectile at the highest point is half of its initial kinetic energy, then what is the angle of projection?

Answer (1 of 3): So the angle is 45° .

If the kinetic energy of a projectile at the highest point is half of its initial kinetic energy, then what is the angle of projection?

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Sort Hamza Said

Former Studying in Giki3y

So the angle is 45° .

Related questions

The kinetic energy of a projectile at its highest point is 75% of its initial value. What is the angle of projection?

A particle is launched at 45 degrees to the horizontal with initial kinetic energy E. Assuming that air resistance to be negligible, what will be the kinetic energy of the projectile when it reaches it's highest point?

What is the angle of a projection for which the kinetic energy at the highest point of its trajectory is equal to one fourth of its kinetic energy at the point of the projection?

What will be the kinetic energy at the highest point if a projectile fired with velocity u at an angle?

What is the ratio of the potential to kinetic energy of a projectile at its highest point if fired at an angle, A, to the horizontal?

Colin Gordon

Teaching Assistant (T.A.) at Michigan State University5y

45 degrees.

At max height all kinetic energy is all in the perpendicular directions of gravity. Thus the parallel must be equal to the perpendicular initially for it to be halfed. That implies the parallel speed is the same as the perpendicular initially. A symmetry only possible if the angle of launch was 45 degrees.

Roy Narten

Mechanical engineer, former engineering instructorAuthor has 1.8K answers and 2.9M answer views3y

Related

If k is the kinetic energy of a projectile fired at an angle of 45 degrees, then what is its kinetic energy at the highest?

The initial kinetic energy is:

k= 1 2 m V 2 k=12mV2

If we assume negligible air resistance then there will be no deceleration in the x-direction. Therefore

V x =Vcosθ Vx=Vcosθ

will remain constant for the entire flight.

When the projectile reaches the highest point, the velocity in the y-direction is zero. So the only kinetic energy it will have is due to its horizontal velocity:

At the highest point,

KE= 1 2 m(Vcosθ ) 2 KE=12m(Vcosθ)2 or KE= 1 2 m V 2 (cos45 ) 2 KE=12mV2(cos45)2 or KE= 1 2 m V 2 (0.5) KE=12mV2(0.5) or KE=0.5k KE=0.5k Sponsored by USAFIS

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Surrinder Nayar

Works at Retirement4y

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A projectile is projected with kinetic energy K. If it has the maximum possible horizontal range, then what will be its kinetic energy at the highest point?

0.5K. Other half is converted to potential energy

James Kewo

Teaching Supervisor (2017–present)Upvoted by

Waqar Ahmad

, Master in physics Physics & Physics, Chemistry, Maths, and Biology, UST Bannu (2015)Author has 66 answers and 193.3K answer views3y

Related

What is the angle of a projection for which the kinetic energy at the highest point of its trajectory is equal to one fourth of its kinetic energy at the point of the projection?

Let’s start with some definition of variables. An object of mass

m m is projected at θ θ

to the horizontal at velocity

v v .

Therefore, initial KE

K E i = 1 2 m v 2 KEi=12mv2

Since the projection is at an angle with the horizontal, we should regard the velocity of the object,

v v

to split into its component perpendicular vectors in the horizontal and vertical directions, namely

v x =vcosθ; v y =vsinθ vx=vcosθ;vy=vsinθ

At the highest point of its trajectory, the velocity is horizontal.

v y =o. vy=o. Therefore, K E f = 1 2 m v 2 x = 1 2 m v 2 cos 2

KEf=12mvx2=12mv2cos2

Since the kinetic energy at t

Roy Narten

Mechanical engineer, former engineering instructorAuthor has 1.8K answers and 2.9M answer views3y

Related

What is the kinetic energy at maximum height when a ball whose kinetic energy is E , is projected at an angle of 45 degrees with vertical motion?

An object projected at an angle has both horizontal motion and vertical motion. Air resistance is commonly ignored, so the horizontal component of the velocity remains constant.

If the ball is projected at

45 ∘ 45∘

, then the initial velocity in each component direction will be the same:

( V i ) x =( V i ) y =Vcos45 (Vi)x=(Vi)y=Vcos45

When the ball reaches the maximum height, the vertical velocity = 0, so the vertical kinetic energy is zero. But the ball is still moving horizontally with velocity =

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