Question

9. A person observes a bird on a tree 39.6 m high and
at a distance of 59.2 m. With what velocity the
person should throw an arrow at an angle of 45° so
that it may hit the bird? ​

Answer 1

Answer:

$\boxed{\mathfrak{Initial \ velocity \ of \ arrow = 41.86 \ m/s}}$

Given:

Horizontal distance (x) = 59.2 m

Vertical distance (y) = 39.6 m

Angle of projection (θ) = 45°

Explanation:

Equation of trajectory:

$\boxed{ \bold{y = xtan \theta - \frac{g {x}^{2} }{2 {u}^{2} {cos}^{2} \theta } }}$

g → Acceleration due to gravity (9.8 m/s²)

u → Initial velocity of projectile (Velocity with which person should throw the arrow so that it may hit the bird)

By substituting values in the equation we get:

$\rm \implies 39.6 = 59.2 \: tan45 \degree - \dfrac{9.8 \times {(59.2)}^{2} }{2 {u}^{2} \: {cos}^{2} 45 \degree} \\ \\ \rm \implies 39.6 = 59.2 \times 1 - \dfrac{9.8 \times 3504.64}{ \cancel{2} {u}^{2} \times \dfrac{1}{ \cancel{2}} } \\ \\ \rm \implies 39.6 = 59.2 - \dfrac{9.8 \times 3504.64}{ {u}^{2} } \\ \\ \rm \implies \dfrac{9.8 \times 3504.64}{ {u}^{2} } = 59.2 - 39.6 \\ \\ \rm \implies \dfrac{9.8 \times 3504.64}{ {u}^{2} } = 19.6 \\ \\ \rm \implies {u}^{2} = \dfrac{9.8 \times 3504.64}{19.6} \\ \\ \rm \implies {u}^{2} = \dfrac{ 3504.64}{2} \\ \\ \rm \implies {u}^{2} = 1752.32 \\ \\ \rm \implies u = \sqrt{1752.32} \\ \\ \rm \implies u = 41.86 \: m {s}^{ - 1}$

$\therefore$ Velocity at which the person should throw an arrow at an angle of 45° so that it may hit the bird = 41.86 m/s