Question

A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball is caught at the same height it was tossed at, how far away is the teammate?​

Answer 1

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

$\boxed{\mathfrak{range = \frac{ u {}^{2} \sin 2\theta }{g} }}$

u is the velocity during its take off and $\theta$ is the angle at which its thrown

Given that

• u = 8m/ s
• $\theta$ = 40°

calculating range using the above formula

$= \frac{ {8}^{2} \sin2(40) }{10}$

$= \frac{64 \times \sin(80) }{10}$

value of sin 80 = 0. 985

$= \frac{64 \times 0.985}{10}$

$= \frac{63.027}{10}$

$= 6.3027$

Hence,

$\mathfrak { \blue{the \: teammate \: is \: \red{\underline{6.3027 \: meters} }\: away } }$