Let's break down the problem and solve each part systematically.

Given Data:

• Initial speed $v_0 = 80 \, \text{m/s}$
• Launch angle $\theta = 30^\circ$
• Height of the building $h = 90 \, \text{m}$
• Acceleration due to gravity $g = 9.8 \, \text{m/s}^2$

We need to calculate: A) The time it takes for the ball to reach the ground.
B) The speed just before it hits the ground.
C) The horizontal distance traveled.

Step 1: Decompose the Initial Velocity

The initial velocity can be decomposed into horizontal ($v_{0x}$) and vertical components ($v_{0y}$):

$v_{0x} = v_0 \cos(\theta) = 80 \cos(30^\circ) = 80 \times \frac{\sqrt{3}}{2} \approx 69.28 \, \text{m/s}$

$v_{0y} = v_0 \sin(\theta) = 80 \sin(30^\circ) = 80 \times \frac{1}{2} = 40 \, \text{m/s}$

Step 2: Calculate Time to Hit the Ground (Part A)

The vertical displacement is governed by the following kinematic equation:

$y(t) = v_{0y} t - \frac{1}{2} g t^2 + h_0$

Where:

• $y(t) = 0$ at the ground.
• $v_{0y} = 40 \, \text{m/s}$
• $g = 9.8 \, \text{m/s}^2$
• $h_0 = 90 \, \text{m}$

Setting the equation for vertical displacement equal to zero to find the total time $t$ until it hits the ground:

$0 = 90 + 40t - \frac{1}{2} (9.8)t^2$

Simplifying the equation:

$4.9t^2 - 40t - 90 = 0$

This is a quadratic equation, and we solve it using the quadratic formula:

$t = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(4.9)(-90)}}{2(4.9)}$

$t = \frac{40 \pm \sqrt{1600 + 1764}}{9.8}$

$t = \frac{40 \pm \sqrt{3364}}{9.8}$

$t = \frac{40 \pm 58}{9.8}$

We take the positive root:

$t = \frac{40 + 58}{9.8} = \frac{98}{9.8} \approx 10 \, \text{seconds}$

So, the time to hit the ground is approximately 10 seconds.

Step 3: Calculate the Speed Just Before Impact (Part B)

The velocity right before hitting the ground consists of both horizontal and vertical components. The horizontal velocity remains constant:

$v_x = v_{0x} = 69.28 \, \text{m/s}$

The vertical velocity changes due to gravity. The vertical velocity at time $t = 10 \, \text{s}$ is:

$v_y = v_{0y} - g t = 40 - 9.8 \times 10 = 40 - 98 = -58 \, \text{m/s}$

Now, we find the magnitude of the total velocity right before impact:

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(69.28)^2 + (-58)^2}$

$v = \sqrt{4809.47 + 3364} = \sqrt{8173.47} \approx 90.41 \, \text{m/s}$

So, the speed just before impact is approximately 90.41 m/s.

Step 4: Calculate the Horizontal Distance (Part C)

The horizontal distance is simply the horizontal velocity multiplied by the total time:

$x = v_{0x} \times t = 69.28 \times 10 \approx 692.8 \, \text{m}$

So, the horizontal distance traveled is approximately 692.8 meters.

Final Answers:

A) Time to hit the ground: $\approx 10 \, \text{seconds}$
B) Speed just before impact: $\approx 90.41 \, \text{m/s}$
C) Horizontal distance: $\approx 692.8 \, \text{m}$

Would you like further details or have any questions?

Here are 5 related questions:

1. How does changing the launch angle affect the time of flight?
2. What would happen if the initial speed was doubled?
3. How is the trajectory affected if air resistance is considered?
4. What is the highest point the ball reaches in this trajectory?
5. How does the initial height affect the range of the projectile?

Tip: Always break down a projectile motion problem into horizontal and vertical components to simplify calculations.