Question

A projectile (with mass 300g) is launched with an initial speed of 30 m/s at an angle of 32 degrees above the horizontal from the edge of a 20 m tall cliff. At the point in the trajectory where the projectile was at its maximum height, a 25g projectile is launched straight up into the first projectile. The smaller mass is traveling at 300 m/s and the two objects stick together. Calculate the percentage increase in height and total horizontal distance traveled for this combined projectile (as compared to what the first projectile would do without being struck by the smaller projectile).

Answer 1

at maximum height, vy = 0

vf = vi + a t

0 = (30 sin32) + (-9.81 t)

t1 = 1.62 sec

horizontal distance covered,

x1 = (v0x) t = (30 cos32)(1.62) = 41.23 m

height above cliff. 0^2 - (30 sin32)^2 = 2(-9.8)(h1 )

h1 = 12.9 m

H = 20 + h1 = 32.9 m  

Applying momentum conservation for the collision,

0 + (25 x 300) = (25 + 300) vy

vy = 23.1 m/s

0^2 - 23.1^2 = (2 x -9.8)(h2)

h2 = 27.2m

%change in height = (h2 / H) x 100

= 82.75 % ................Ans

if not stuck then,

time to strike ground,

0 - 20 = (30 sin32)(t) - 9.8 t^2 / 2

4.9 t^2 - 15.9t - 20 = 0

t = 4.22 s

X = (v0x) t = 107.4 m

if struck then,

0 - (32.9) = 23.1t - 9.8 t^2 /2

4.9 t^2 - 23.1t - 32.9 = 0

t = 5.86 s

after collision, (300 x 30 cos32) + (25 x 0) = (300 + 25)vx

vx = 23.5 m/s

x2 = (23.5)(5.86) = 137.6 m

X' = 41.23 + 137.6 = 178.85 m

% change in horizontal distance,

= ((X' - X)/X)(100)

= ((178.85 - 107.4)/(107.4)) x 100

= 66.5 %