Question

The maximum height reached by a projectile is equal to half of the range of projectile. If the initial velocity of projection was 20 m/s, then what was maximum height reached? (g = 10 m/S2)​

Answer 1

Answer:

• We know a relation

• R tanθ = 4 H

Given: R = 2H ; u = 20m/s

• That implies, H/R = tanθ/4

⇒ H/2H = tanθ/4

• 1/2 = tanθ /4

We'll get tanθ= 2

Now we'll draw a triangle to calculate all trignometric frnctions easily

You can see the triangle in the Attachment.

From the triangle sinθ = $\frac{2}{\sqrt[]{5} }$ ; cosθ = $\frac{1}{\sqrt[]{5} }$

H = $\frac{u^2sin^2theta}{2g}$

H = $\frac{400 * (\frac{2}{\sqrt{5} })^2 }{20} = \frac{4}{5} *20 = 16m$

H = 16m

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Answer 2

Given:

Initial velocity $=20m/s$

Acceleration due to gravity $=10m/s^{2}$

To find:

Maximum height reached by the projectile.

Solution:

Step 1

For a projectile released from a point with some initial velocity as $u$ and the angle made by the projectile with the horizontal,

The height reached by the projectile $H=\frac{u^{2} .sin^{2}\alpha }{2g}$  

The range of the projectile is $R=\frac{u^{2}.sin 2\alpha }{g}$  

Step 2

Now,

According to the question,

The height reached by the projectile is equal to half the range of the projectile. Hence, we get

$\frac{u^{2} .sin^{2}\alpha }{2g}=\frac{1}{2}\frac{u^{2}.sin 2\alpha }{g}$

$sin^{2} \alpha =sin2\alpha$

$sin^{2} \alpha =2.sin\alpha .cos\alpha$

$sin\alpha =2.cos\alpha$

$tan\alpha =2$

We know,

$sec^{2} \alpha =1+tan^{2} \alpha$

$sec^{2} \alpha =5$

Hence,

$cos^{2} \alpha =\frac{1}{5}$

We know,

$sin^{2} \alpha =1-cos^{2} \alpha$

$sin^{2} \alpha =\frac{4}{5}$

Step 3

Now,

We have, $u=20m/s$   $;$ $sin^{2} \alpha =\frac{4}{5}$    $;$ $g=10m/s^{2}$

We get, the height reached by the projectile as

$H=\frac{u^{2} .sin^{2}\alpha }{2g}$

$H=\frac{(20)^{2}(\frac{4}{5} ) }{2(10)}$

$H=16m$

Final answer:

Hence, the maximum height reached by the projectile is 16 meter.