Question

A ball is projected at an angle of 45° to the horizontal. If the range is 10 m, the maximum
m (g = 10 m/s
height attained by the ball is
​

Answer 1

Answer -

Given -

$\longrightarrow$$\theta$ = 45°

$\longrightarrow$R = 10m

where

$\longrightarrow$$\theta$ is the angle of projection.

$\longrightarrow$R is range.

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To find -

Maximum height $\longrightarrow$ H

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Formula used -

$\sf{R = \frac{ {u}^{2} \sin2 \theta }{g} }$

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

where

$\longrightarrow$R is range.

$\longrightarrow$u is initial velocity.

$\longrightarrow$g is acceleration due to gravity.

$\longrightarrow$H is maximum height.

$\longrightarrow$$\theta$ is the angle of projection.

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Solution

Substituting the value in

$\sf{R = \frac{ {u}^{2} \sin2 \theta }{g} }$

$\implies$$\sf{10 = \frac{{u}^{2} \sin2(45) </p><p>}{10} }$

$\implies$$\sf{10 = \frac{ {u}^{2} }{10}}$

$\implies$$\sf{{u}^{2} = 10 \times 10}$

$\implies$$\sf{u = 10 m/s }$

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$\longrightarrow$R = 10 m

$\longrightarrow$u = 10 m/s

$\longrightarrow$$\theta$ = 45°

Substituting the value in

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

$\implies$$\sf{H = \frac{ {10}^{2} \times \sin^{2} 45}{2 \times 10}}$

$\implies$$\sf{\frac{100}{20} }$

$\implies$$\sf{H = 5m }$

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Thanks

Answer 2

Dear student,

It is given that the ball is projected at an angle of 45° to the horizontal. Also, the horizontal range of the projectile is 10 m. We need to find the maximum height attained by the ball.

We know by the concepts of projectile motion and motion in two dimension, that, horizontal range is given by;

                    R = u²sin(2∅)/ g

Putting the given values in the above formula, we get;

             10 = u² sin (45 × 2)/ 10

           100 = u² sin 90°

           100 = u² × 1

              u = √100

             u = 10 m

Now, we know, that for maximum height;

               H = u²/ 2g

Putting the given and obtained values, we get;

        H = 10²/ 2 × 10

        H = 100 / 2 × 10

        H = 5 m            

Hence, the maximum height attained by the projectile is 5 m.