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

$v_o = 7.76 m/s$

Answer 2

The magnitude of the balloon's initial velocity is 7.80 m/s

The given parameter

• height from which the balloon was thrown, h = 1.82 m

• the initial velocity of the balloon, = v₀

• the angle of projection, θ = 29⁰

• the horizontal distance of the balloon, x = 7.5 m

To find:

• the magnitude of the balloon's initial velocity

The vertical component of the velocity is calculated as;

$v_y = v_0\times sin(\theta)\\\\v_y = v_0 \times sin(29)\\\\v_y = 0.4848v_0$

The horizontal component of the velocity is given as;

$v_x = v_0\times cos(\theta)\\\\v_x = v_0 \times cos(29)\\\\v_x = 0.8746v_0$

The horizontal distance or range of the projectile is given as;

$X = v_xt \\\\7.5 = 0.8746v_0t\\\\v_0t = \frac{7.5}{0.8746} \\\\v_ot = 8.575$

The change in vertical height of the balloon is given as;

$\Delta h = v_yt - \frac{1}{2}gt^2\\\\h_2-h_1 = v_yt - \frac{1}{2}gt^2$

assume the balloon falls to zero position

$0-1.82 = v_yt - \frac{1}{2} gt^2\\\\-1.82 = v_yt - \frac{1}{2} gt^2\\\\-1.82 = 0.4848v_0t - \frac{1}{2} (9.8)t^2\\\\-1.82 = 0.4848(8.575) - 4.9t^2\\\\4.9t^2 = 4.157 + 1.82\\\\4.9t^2 = 5.977\\\\t^2 = \frac{5.977}{4.9} \\\\t^2 = 1.2198\\\\t = \sqrt{1.2198} \\\\t = 1.1 \ s$

The magnitude of initial velocity of the balloon is calculated as;

$v_0t = 8.575\\\\1.1v_0 = 8.575\\\\v_0 = \frac{8.575}{1.1} \\\\v_0 = 7.80 \ m/s$

Thus, the magnitude of the balloon's initial velocity is 7.80 m/s

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