Question

A water balloon launcher (a is used to propel a water balloon from a sidewalk that is 48 ft above a level parking lot on campus. The water balloon leaves the launcher at a speed of 64 ft/sec at an angle of 30 degrees with the horizontal. a) When will the water balloon hit the pavement? b) How far will the balloon be horizontally from the launch point at that time?

Answer 1

a.)

Using second kinematics in vertical direction,

Sy = Uy*t + 0.5*ay*t^2

here, Sy = vertical height = -48 ft (As, pavement is below the initial position.)

Uy = initial vertical speed = U*sin

U = 64 ft/sec.

= projection angle = 30 deg

ay = acceleration in vertical direction = -g = -32.2 ft/s^2

t = time required to hit the pavement = ??

So,

-48 = 64*sin(30 deg)*t + 0.5*(-32.2)*t^2

16.1*t^2 - 32*t - 48 = 0

By solving above quadratic equation,

t = [32 sqrt(32^2 + 4*16.1*48)]/(2*16.1)

Possible value of time will be:

t = [32 + sqrt(32^2 + 4*16.1*48)]/(2*16.1)

t = 2.986 s

b.)  

Now, again using second kinematics law in horizontal direction,

Sx = Ux*t + 0.5*ax*t^2

here, Sx = Horizontally distance from launching point = ??

Ux = initial horizontal speed = U*cos = 64*cos(30 deg)

ax = acceleration in horizontal direction = 0

So, at time 't',

Sx = 64*cos(30 deg)*2.986

Sx = 165.5 ft

"Let me know if you have any query."