Question

0. A cricketer can throw a ball to maximum horizontal distance of 125m, calculate maximum
heightimoved by ball. g=10ms-? (31.25m)
​

Answer 1

Answer:

Given:

Max range =125 m

To find:

Max height by the ball.

Calculation:

For Max range, the projectile has to be projected at an angle of 45° with the horizontal.

Range = u²sin(2θ)/g

=> R = u² sin (90°)/g

=> R = u²/g = 125 .......[given]

Now, Max height

= u² sin²(θ)/2g

= u² sin²(45°)/2g

= u² (1/√2)²/2g

= u²/4g

= ¼(u²/g)

= ¼ R

= ¼ (125)

= 31.25 metres.

So max height = 31.25 metres.

Answer 2

Answer:

$\large \boxed{ \sf{31.25 \: m}}$

Explanation:

Clearly, it's given that A cricketer can throw a ball to maximum horizontal distance of 125 m.

• We know that, Maximum range is reached if ball is thrown at 45° from horizontal.

Therefore, We have the condition,

• Range, R = 125 m
• Angle, $\alpha$= 45°
• Maximum Height, H = ??

We know the formula,

• $\large \boxed{ \sf{R= \frac{ {u}^{2} \sin(2 \alpha ) }{g} }}$

Substituting the values, We get,

$\sf{ = > R = \frac{ {u}^{2} \sin(90) }{g}} \\ \\ \sf{= > R = \frac{ {u}^{2} }{g} }\\ \\ \sf{ = > \frac{ {u}^{2} }{g} = 125 \: \: \: \: \: \: \: ..........(1)}$

Further, We know that,

• $\large \boxed{ \sf{H = \frac{ {u}^{2} { \sin }^{2} \alpha }{2g} }}$

Substituting the values, We get

$\sf{= > H= \frac{ {u}^{2} { \sin}^{2}45 }{2g} } \\ \\ \sf{= > H= \frac{ {u}^{2} \times {( \frac{1}{ \sqrt{2} }) }^{2} }{2g} }\\ \\ \sf{ = > H = \frac{1}{4} \times \frac{ {u}^{2} }{g} }\\ \\ \sf{ = > H = \frac{1}{4} \times 125 \: \: \: \: \: \: \: \: \: (from \: eqn. \: 1)}\\ \\ = > \sf{H = 31.25 \: m}$