To determine the maximum height attained by the ball, we can use the equations of projectile motion. Given the initial height of the ball is same as the height of the cricketers (both are $$2.5 \mathrm{~m}$$ tall), the angle of projection is $$30^{\circ}$$, and the horizontal range is $$50 \mathrm{~m}$$, we need to find out the maximum height.

First, let's calculate the initial velocity of the ball. For horizontal range, we use the formula:

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

where:

R is the range = $$50 \mathrm{~m}$$
u is the initial velocity
$$\theta$$ is the angle of projection = $$30^{\circ}$$
g is the acceleration due to gravity = $$9.8 \mathrm{~m/s^2}$$

First, calculate $$\sin 2\theta$$:

$$ \sin 2 \theta = \sin 60^{\circ} = \frac{\sqrt{3}}{2} $$

Thus, the range equation becomes:

$$ 50 = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{9.8} $$

Solve for the initial velocity $$u:$$

$$ u^2 = \frac{50 \cdot 9.8 \cdot 2}{\sqrt{3}} $$

$$ u^2 = \frac{980}{\sqrt{3}} $$

$$ u \approx \sqrt{565.69} $$

$$ u \approx 23.78 \mathrm{~m/s} $$

Now, we need to find the maximum height achieved by the ball. The formula for maximum height in projectile motion is:

$$ H = h_0 + \frac{u^2 \sin^2 \theta}{2g} $$

where:

$$ h_0 $$ is the initial height = $$2.5 \mathrm{~m}$$
$$\sin \theta$$ is for $$\theta = 30^{\circ}$$

First, calculate $$\sin 30^{\circ}$$:

$$ \sin 30^{\circ} = 0.5 $$

Thus, substituting the values, we get:

$$ H = 2.5 + \frac{23.78^2 \cdot 0.5^2}{2 \cdot 9.8} $$

$$ H = 2.5 + \frac{23.78^2 \cdot 0.25}{19.6} $$

$$ H = 2.5 + \frac{565.69 \cdot 0.25}{19.6} $$

$$ H = 2.5 + \frac{141.42}{19.6} $$

$$ H \approx 2.5 + 7.2 $$

$$ H \approx 9.7 \mathrm{~m} $$

Therefore, the maximum height attained by the ball is closest to option D, which is:

Option D: 9.7 m

Motion

Continue this paper

Similar Motion PYQs

More practice from this question