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

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`

Step-by-step 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 + (1)/(2)at^2`

`-1.78 = (v_osin25) t - (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`