Question

a projectile is thrown with a velocity of 10ms-1 at an angle of 30° with horizontal . the value of maximum height gained by it is a. 1 m b. 1.25 m c. 2 m d. 2.5 m​

Answer 1

Explanation:

If you project an object from ground level at 45 degrees to the horizontal the maximum range is…

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I am not using g = 9.8 or whatever because:

(a) you mention “throwing” it. This depends on how tall you are.

This makes it a completely different problem!

In this case the value of R will be greater than 10m

(b) you did not mention whether or not the ground is horizontal.

(c) you did not mention whether or not the object would be affected by air resistance.

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I decided to do a graphical simulation of a cricket ball projected at a 45 degree angle at a velocity of 10 m/s from 3 common heights.

Here I used g = 9.8

hope it will help you

please mark me as brainlist

Answer 2

Given:

Initial velocity (u) = 10m/s.

Angle of projection = 30°.

To Find:

Maximum height of projectile.

Solution:

We know that,

\star \: \boxed{\rm\red{Height \: of \: projectile = \frac{ {u}^{2} {sin \theta}^{2} }{2g}}}⋆Heightofprojectile=2gu2sinθ2

Substituting values,

:\implies\mathtt{ \frac{( {10)}^{2} \times ( {sin30)}^{2} }{2 \times 9.8} }:⟹2×9.8(10)2×(sin30)2

:\implies\mathtt{ \frac{100 \times 0.5 \times 0.5}{19.6} }:⟹19.6100×0.5×0.5

:\implies\mathtt{ \frac{100 \times 0.25}{19.6} }:⟹19.6100×0.25

:\implies\mathtt{\dfrac{\cancel{25}}{\cancel{19.6}}}:⟹19.625

\sf\star \: \underbrace\green{Height \: of \: projectile \: = \: 1.275m} \: \star⋆Heightofprojectile=1.275m⋆

Hence,

Value of maximum height gained is 1.275m.