In order to find the components of the velocity, assume x-axis is horizontal and directed from Caitlin to the wall, and y-axis is vertical and directed up. Then both x and y components of velocity are positive.

The horizontal component of velocity is

`V_x = Vcos(theta)` , where `theta = 15` degrees is...

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In order to find the components of the velocity, assume x-axis is horizontal and directed from Caitlin to the wall, and y-axis is vertical and directed up. Then both x and y components of velocity are positive.

The horizontal component of velocity is

`V_x = Vcos(theta)` , where `theta = 15` degrees is the angle the velocity vector makes with the horizontal. Since V = 20 m/s,

`V_x = 20cos(15) = 19.32` m/s.

The vertical component of velocity is

`V_y = Vsin(theta) = 20sin(15) =5.18` m/s.

Since in the horizontal direction there is no acceleration (the only force present is gravity and it is vertical), the horizontal component of velocity will remain constant. Therefore the horizontal distance traveled by the ball until it hit the wall, d = 6 m, is

`d= V_xt` , where t is the time until the ball hit the wall.

From here, this time can be found:

`t=d/V_x = 6/19.32 = 0.31` seconds.

In the verical direction, the height H at which the ball hit the wall is

`H = h +V_yt - (g*t^2)/2` , where h = 0.5 m is the original height of the ball above the ground, t = 0.31 seconds is the time until the ball hits the wall. There is a negative sign in front of g because the gravity is directed downward.

Plugging in the values for `V_y`

and t, we can find H:

`H=0.5 + 5.18*0.31 - (9.8*(0.31)^2)/2 = ` 1.63 m.

**a) The ball is 1.63 m above the ground when it hits the wall.**

The vertical component of velocity of the ball when it hits the wall, `V_(yf)` can be found as

`V_(yf) =V_y- g*t = 5.18 - 9.8*0.31 = ` 2.142 m/s

The vertical component of velocity is still positive when it hits the wall, so

**the ball is still on its way up.**

**Further Reading**