Question

A projectile is launched from ground level with an initial speed of 53.2 m/s. Find the launch angle (the angle the initial velocity vector is above the horizontal) such that the maximum height of the projectile is equal to its horizontal range. (Ignore any effects due to air resistance.)

Answer 1

Answer :-

$\implies \boxed{ \sf{ \theta = { \tan}^{ - 1} (4 \degree)}} \:$

To Find :-

→ inclination angle of launch .

Explanation :-

Given that

• Initial speed (u) = 53.2 m/s

Here no need of find range and height .

According to the question ,its range become equal to its height .

We know that the R by H rule

$\implies \sf{ \frac{R}{H} \: = \frac{4}{ \tan \theta \: } } \\ \\ \because \: \: \sf{R = H\: } \\ \therefore \: \\ \implies \sf{ \frac{R}{R} = \frac{4}{ \tan \theta} } \\ \\ \implies \sf{ \tan \theta = 4 \: } \\ \\ \implies \boxed{ \sf{ \theta = { \tan}^{ - 1} (4 \degree)}}$

And ,if you don't know the R/H rule so you can take place formula.

We know that .

u → initial velocity

$\to \boxed{ \sf{R \: (range) = \frac{ {u}^{2} \sin2 \theta }{g} }} \\ \sf{and \: } \\ \: \\ \to \boxed{ \sf{ H_{max.} = \frac{ {u}^{2} { \sin}^{2} \theta }{2g} }}$

according to the question

$\to \sf{ H_{max.} = R \: } \\ \\ \to \sf{ \frac{ {u}^{2} { \sin}^{2} \theta}{2g} = \frac{ {u}^{2} \sin2 \theta }{g} } \\ \\ \to \: \frac{ { \sin}^{2} \theta }{2} = 2 \sin \theta \cos \theta \\ \\ \to \: \tan \theta = 4 \\ \\ \implies \boxed{ \sf{ \theta = { \tan}^{ - 1} (4 \degree)}} \:$

Answer 2

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QUESTION :

A projectile is launched from ground level with an initial speed of 53.2 m/s. Find the launch angle (the angle the initial velocity vector is above the horizontal) such that the maximum height of the projectile is equal to its horizontal range. (Ignore any effects due to air resistance.)

SOLUTION :

See the above attachment.

Here, we have the following condition :

The maximum height of the projectile is equal to its horizontal range.

So,

We know that :

The range of any projectile is given by :

R = u ^ 2 { sin 2 Phi } / 2 g

Maximum height of any projectile is equal to :

H _ { Max } = u ^ 2 { Sin Phi } ^ 2 / g

R = H _ { Max }

=> u ^ 2 { sin 2 Phi } / 2 g = u ^ 2 { Sin Phi } ^ 2 / g

Sin 2 Phi = 2 sin phi cos phi

Now , cancelling U ^ 2 and sin theta :

=> Tan theta = 4 °

Theta is tan ^ -1 { 4 ° }

ANSWER :

The answer is tan ^ -1 { 4 ° }