Question

A student throws a water balloon with speed v0 from a height h = 1.82 m at an angle θ = 29° above the horizontal toward a target on the ground. The target is located a horizontal distance d = 7.5 m from the student’s feet. Assume that the balloon moves without air resistance. Use a Cartesian coordinate system with the origin at the balloon's initial position. Determine the magnitude of the balloon's initial velocity,v0, in meters per second.

Answer 1

Answer:

$v_o = 7.76 m/s$

Explanation:

As we projected the balloon at speed vo at an angle of 29 degree

so the two component of velocity is given as

$v_x = v_ocos29 = 0.875 v_o$

$v_y = v_o sin29 = 0.485 v_o$

now we know that in x direction we have

$d = v_x t$

$7.5 = 0.875 v_o t$

$v_o t = 8.57$

in y direction we have

$- 1.82 = (0.485 v_o) t - \frac{1}{2}gt^2$

$-1.82 = 0.485(8.57) - 4.9 t^2$

$t = 1.1 s$

now we have

$v_o = 7.76 m/s$