Question

An arrow is shot at an angle of 40° to the horizontal with a velocity 19.6 m/s. Find maximum height, time of flight, and horizontal range.

Answer 1

Answer:

Time of flight of projectile is given by formula,

T = (2 × u × sin(theta))/g

Where u is initial velocity, g is acceleration due to gravity, and theta is angle made by initial velocity with horizontal.

Given in the problem, theta = 30 degree, initial velocity is 19.6 m/s and g =9.8 m/s^2.

T = (2 × 19.6 × (1/2))/9.8

T = 19.6/9.8 = 2

T = 2 seconds.

Hope this will help

Answer 2

Given: Horizontal velocity = 19.6 ms⁻¹

           Angle = 45°

To find: (a) maximum height

            (b) time of flight

            (c) horizontal range

Solution: According to the projectile motion,

• The maximum height of the projectile is when that projectile reaches zero vertical velocity and is denoted by u²Sin²θ/2g

• Time of flight is the time taken by the object to cover the whole projectile and is denoted by 2uSinθ/g

• Horizontal range = u²Sin2θ/g

(a) Maximum height = (19.6)² Sin²45°/2×9.8

                                = 19.6/2

                                = 9.8 m

(b) Time of flight = 2uSinθ/g

                           = 2× 19.6Sin45°/9.8

                           = 2√2 seconds

(c) Horizontal range = u²Sin2θ/g

                                 = (19.6)²Sin2 (45°)/9.8

                                 = 19.6 ×2 = 39.2 m

Therefore, maximum height = 9.8m

                , time of flight = 2√2 seconds

                , horizontal range = 39.2 m