A ball is tossed from an upper-storey window of a building.The ball is given an initial velocity of 8 m/s at an angle of 20 degrees below the horizontal.it strikes the ground 3s later .(a) How far horizontally from the base of the building does the ball strike the ground? (b)Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0m below the level of launching?

Vo = 8 m/s. @ 20o = Initial velocity.

Xo = 8*cos20 = 7.52 m/s = Hor. comp.
Yo = 8*sin20 = 2.74 m/s = Ver. comp.

a. d = Xo * Tf = 7.52m/s * 3s.=22.56 m.

b. h = Yo*Tf + 0.5g*Tf^2.
h = 2.74*3 + 4.9*3^2 = 52.32 m.

c. h = Yo*t + 0.5g*t^2 = 10 m.
2.74t + 4.9t^2 = 10
4.9t^2 + 2.74t - 10 = 0
Use Quad. Formula:
t = 1.18 s.

how did you get that 0.5g??

in reply to "how did you get that 0.5g??":

0.5g= (1/2)(g)=(1/2)(9.8)

To solve this problem, we need to break it down into two components: the horizontal and vertical motion of the ball.

Let's start with the horizontal component:

(a) How far horizontally does the ball strike the ground?

In horizontal motion, there is no acceleration acting on the ball. The initial horizontal velocity is given as 8 m/s and remains constant throughout the motion.

To find the horizontal distance traveled by the ball, we can use the formula:

Distance = Velocity × Time

Given that the time of flight is 3 seconds, and the initial horizontal velocity is 8 m/s, we can calculate the horizontal distance traveled:

Distance = 8 m/s × 3 s = 24 meters

Therefore, the ball strikes the ground 24 meters horizontally from the base of the building.

Now, let's move on to the vertical component:

(b) Find the height from which the ball was thrown.

In the vertical motion, we need to consider two things: the initial vertical velocity (upwards) and the acceleration due to gravity.

The initial vertical velocity can be found by considering the given angle below the horizontal. We can break down the initial velocity into its vertical and horizontal components using trigonometry.

Vertical component: V_y = V_initial * sin(angle)
V_y = 8 m/s * sin(20 degrees)
V_y ≈ 2.711 m/s (rounding to three decimal places)

Now, we can use the kinematic equation to find the height from which the ball was thrown:

Height = V_initial^2 * sin^2(angle) / (2 * g)

Where g is the acceleration due to gravity (9.8 m/s^2).

Height = (8 m/s)^2 * (sin(20 degrees))^2 / (2 * 9.8 m/s^2)
Height ≈ 0.235 meters (rounding to three decimal places)

Therefore, the ball was thrown from a height of approximately 0.235 meters.

Moving on to the last part:

(c) How long does it take the ball to reach a point 10.0m below the level of launching?

We need to calculate the time it takes for the ball to reach a height below the launching point. Let's call this time t.

Using the equation for the vertical motion:

Height = V_initial * t + (1/2) * g * t^2

Given that the initial vertical velocity was 2.711 m/s, height is -10.0 m (negative because it's below the launching point), and g is 9.8 m/s^2, we can solve for t.

-10.0 m = 2.711 m/s * t + (1/2) * 9.8 m/s^2 * t^2

Simplifying the equation and rearranging it:

4.9 t^2 + 2.711 t - 10.0 = 0

We can solve this quadratic equation using various methods, such as factoring, completing the square, or using the quadratic formula. Let's solve it using the quadratic formula:

t = (-2.711 ± √(2.711^2 - 4 * 4.9 * -10.0)) / (2 * 4.9)

t ≈ 0.586 seconds (rounding to three decimal places)

Therefore, it takes approximately 0.586 seconds for the ball to reach a point 10.0 meters below the level of launching.