Question

A projectile is launched with an initial speed of 40 m/s at an angle of 25° above the horizontal.

(a) What are the horizontal and the vertical components of initial velocity.

(b) Find the time taken by the projectile to reach the highest point and its height at the highest point.

(c) How long does it take the projectile to hit the ground after launch and how far from the starting point it hits the ground.

(d) Calculate the velocity (magnitude and the angle) of the projectile 3.0 s after launch.

(e) What is the position (horizontal and vertical) of the projectile 3.0 s after launch?

Answer 1

given that

initial speed u = 40 m/s

angle = 25 degrees

a) x component of initial velocity ux = ucos(theta) = 40 cos 25 = 36.25 m/s

y component of velocity uy = usin(theta) = 40 sin(25) = 16.9 m/s

b) time taken to reach the highest point t = usin(theta)/g

t = 16.9/9.81 = 1.72 sec

maximum height h max = u^2*sin^2(theta)/2g

h max = 16.9^2/2*9.81 = 14.6 m

c) total time T = 2usin(theta)/g

T = 2*16.9/9.81 = 3.44 sec

total horizontal distance = u^2*sin(2theta)/g = 40^2*sin(2*25)/9.81 = 124.94 m

d) after 3 sec

vx = ux = 36.25 m/s

vy = uy-gt

vy = 16.9-9.81*3 = -12.53 m/s

magnitude v = sqrt(vx^2+vy^2)

v = 38.35 m/s

angle = tan^-1(-12.53/36.25) = 19.06 degrees below the x axis

e) after 3 sec

horizontal distance x = ux*t = 36.25*3 = 108.75 m

vertical distance y = uy*t-1/2*gt^2

y = 16.9*3-1/2*9.81*3^2

y = 50.7-44.145

y = 6.555 m