Question

A water balloon is tossed at an angle ϕ = 37° above the horizontal, from a height of 1.50-m above the ground to a target located Δx = 13.0 m away. The target is located on a platform h = 3.4 m above the ground. What must the initial speed of the water balloon be to hit the target? answer in m/s

Answer 1

let, the initial velocity is v m/s. The time to hit the target is t sec.

From one dimensional motion:

y = y0 + v(sin37).t - (0.5)×g×t²

here, y = vertical position of the balloon after t sec= 3.4 m

y0 = initial vertical position of the balloon= 1.5 m

g = gravitational acceleration= 9.8 m/s²

Now, 3.4 = 1.5 + 0.6018(v.t) - 4.9t²

=> 4.9(t²) - 0.6018(v.t) + 1.9 = 0..........i)

Again, v(cos37)×t = 13

=> v.t = 16.28

putting the value of (v.t) in the equation i) :

4.9×t² - 0.6018×(16.28) + 1.9 =0

=> 4.9t² -9.796 + 1.9= 0

=> 4.9t² = 7.896

=> t² = 1.61

=> t = 1.27 sec (approx)

The initial velocity of the balloon is= (16.28/1.27)= 12.82 m/s

Please comment if you have any doubt and like if it helps.

Happy learning.