Question

A student throws a water balloon with speed v0 from a height h = 1.78 m at an angle θ = 25° 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.

Answer 1

Answer:

time after which it will hit the floor

t = 1.038 s

initial speed is given as

$v_o = 7.97 m/s$

final component of velocities

$v_x = 7.23 m/s$

$v_y = -6.81 m/s$

Explanation:

Height of the water balloon from which it is projected is given as

h = 1.78 m

now we know that

$\Delta y = v_y t + \frac{1}{2}at^2$

$-1.78 = (v_osin25) t - \frac{1}{2}(9.81)t^2$

also we know that

$v_o cos25 t = 7.5$

now we have

$-1.78 = (7.5 tan25) - 4.9 t^2$

now we have

$t = 1.038 s$

Now we have

$v_o cos25 (1.038) = 7.5$

$v_o = 7.97 m/s$

Now final speed when it hit the floor in x and y direction is given as

$v_x = v_o cos25 = 7.23 m/s$

$v_y = v_o sin25 - (9.81)(1.038)$

$v_y = 7.97 sin25 - 10.18$

$v_y = -6.81 m/s$