Question

a A boy can throw a stone up to a maximum height of 1 m. The maximum horizontal distance that the boy can throw the same stone is up to x m. Find value of x.​

Answer 1

I really don't know this

Answer 2

Given: A boy can throw a stone up to a maximum height of 1 m.

To find: Maximum horizontal distance that the boy can throw the stone

Explanation: When the boy throws stone vertically, acceleration due to gravity(g)= 10 m/s^2 and final velocity at maximum height(v)= 0. Let his throwing speed be u and s=1 m be distance

Using kinematics equation,

${v}^{2} - {u}^{2} = - 2as$

${u}^{2} = 2 \times 10 \times 1$

${u}^{2} = 20$

Now, when stone is thrown horizontally the maximum distance can that be covered is when ball is thrown at angle 45°. Therefore, the formula used for range of projectile when angle= 45° is:

$distance = \frac{ {u}^{2} \sin(2 \times 45) }{g}$

${u}^{2} = 20(found \: before)$

= 20 * 1 / 10 ( sin 90°=1)

= 2 m

Therefore, the maximum horizontal distance the boy can throw the stone is 2 m.