Question

An elastic loaded balloon launcher fires balloons at an angle of [38.0o above horizontal] from the surface of the ground. if the initial velocity is 25.0 m/s, find how far away the balloons are from the launcher when they hit the level ground again. 9)a movie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. how fast must the motorcycle leave the cliff-top if it's to land on the level ground below at a distance of 90.0 m from the base of the cliff?

Answer 1

#1

a balloon is launched at an angle 38 degree from the horizontal with speed 25 m/s

now the two component of the velocity of balloon is given as

$v_x = vcos38$

$v_x = 25 cos38 = 19.7 m/s$

$v_y = vsin38$

$v_y = 25 sin38 = 15.4 m/s$

Now the time of flight of the motion of balloon is given as

$T = \frac{2v_y}{g}$

$T = \frac{2*15.4}{9.8}$

$T = 3.14 s$

now the horizontal distance traveled by the balloon in the given time is

$R = v_x * T$

$R = 19.7* 3.14 = 62 m$

so the balloon will land at a distance of 62 m

#9

Motor cycle moves off horizontally with speed "v"

now the height of the cliff is given as

$h = 50 m$

now time taken by the motor cycle to reach the bottom

$y = \frac{1}{2}gt^2$

$50 = \frac{1}{2}*9.8*t^2$

$t = 3.2 s$

now we know that in the given time motorcycle will cover horizontal distance 90 m

so we can say

$v* t = 90$

$v*3.2 = 90$

$v = 28.2 m/s$

so motorcycle will move off with speed 28.2 m/s