Question

An arrow is shot at an angle 38 ◦ with the horizontal. It has a velocity of 46 m/s. How high will the arrow go?

Answer 1

Answer:

40·919 m

Explanation:

Initial velocity of the arrow = 46 m/s

Angle at which it is thrown from horizontal = 38°

At the maximum height, the vertical component of velocity will be 0

Initial velocity in vertical direction = 46 × sin(38) = 28·32 m/s

From the formula

v² - u² = 2 × a × s

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Considering the formula in vertical direction and taking upward direction as positive

v = 0

u = 28·32 m/s

a = - g = - 9·8 m/s²

Let s be the maximum height

- 28·32² = - 2 × 9.8 × s

⇒ s = 40·919 m

∴ The arrow will go 40·919 m high