Question

An arrow strikes the target at the end of a 27-meter-long lane. It lands at an angle of 4 degrees below the horizontal. What is the initial velocity at which the arrow was launched? Assume that the initial velocity toward the target is entirely horizontal and the angel of the arrow in the target is the direction of the final velocity.

Answer 1

Given that Range of arrow is 27.0 m

And Arrow is making an angle 4.00 degree with the ground after it stuck into the ground

Now We know that in projectile motion horizontal velocity remains constant, Since there is no acceleration in horizontal direction and here given that projectile is shot parallel to the ground, So

Suppose final velocity of arrow is

V = Vx i + Vy j

Where |V| = sqrt (Vx^2 + Vy^2)

\theta = arctan (Vy/Vx) = 4.00 deg (Given)

Vy/Vx = tan

Vy = Vx*tan

Now Since horizontal velocity is always constant and projectile is shot parallel to the ground, So

V0 = Initial velocity = V0x = Vx

V0y = 0, since shot parallel to the ground

Now Since no acceleration in horizontal direction, So

R = V0x*t = 27.0 m

t = 27.0/V0x = 27.0/Vx

Now Using 1st kinematic equation in vertical direction:

Vy = V0y + a*t

a = acceleration of arrow in vertical direction = g = 9.81 m/sec^2

Vy = 0 + g*t

(Since Vy = Vx*tan , g = 9.81 m/sec^2, t = 27.0/Vx), So Using these values:

Vx*tan = g*(R/Vx)

Vx^2 = g*R/tan

Vx = sqrt (9.81*27.0/tan 4.00 deg)

Vx = 61.55 m/sec

Initial Speed of arrow = 61.55 m/sec

(If need in 2 sig fig, than use 62 m/sec, if need in 3 sig fig then use 61.6 m/sec)

Let me know if you've any query,