Question

Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only force acting on it thereafter is gravity. Take \(g=32 \mathrm{ft} / \mathrm{sec}^{2}\). (a) Find the highest altitude attained by the object. (b) Determine how long it takes for the object to fall to the ground.

Short answer

Answer: The highest altitude attained by the object is 576 feet, and it takes approximately 2.83 seconds for the object to fall to the ground.

Step-by-step solution

Identify Variables and Constants from the Given Exercise

The variables and constants are as follows: - Initial height (s0) = 320 feet - Initial velocity (v0) = 128 ft/sec (upward) - Gravity (g) = 32 ft/sec² (downward) - Time (t) For part (a), we need to find the maximum height (highest altitude), and for part (b), we need to find the time it takes for the object to fall to the ground.

Find Maximum Height (Part A)

To find the maximum height (highest altitude), we first find the time when the object reaches the maximum height. This happens when the vertical velocity is zero. We use the Velocity equation: \(v(t) = v_0 - gt\) Setting v(t) = 0, we solve for time t: \(0 = 128 - 32t\) \(t = \frac{128}{32} = 4\) Now, we can find the maximum height using the Position equation: \(s(t) = s_0 + v_0t - 0.5gt^2\) Plugging the given values: \(s(4) = 320 + 128(4) - 0.5(32)(4^2)\) \(s(4) = 320 + 512 - 256 = 576\) Therefore, the highest altitude attained by the object is 576 feet.

Find the Time It Takes for the Object to Fall to the Ground (Part B)

To find the time it takes for the object to fall to the ground, we need to find the time when the object's height is zero (assuming the ground is at zero height). We use the Position equation and set s(t) = 0: \(0 = 320 + 128t - 0.5(32)t^2\) Rearrange the equation: \(16t^2 - 128t + 320 = 0\) Now, we can solve the quadratic equation by either factoring or using the quadratic formula. Here, we will use the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) \(t = \frac{128 \pm \sqrt{128^2 - 4(16)(320)}}{32}\) After calculating, we get two possible values for t: \(t \approx 2.83\) or \(t \approx 1.17\) Since the object was launched at t = 0, the solution t = 1.17 corresponds to the time when the object reaches the ground on its way up, which is not what we are looking for. Therefore, we discard this solution and consider t = 2.83. So, it takes approximately 2.83 seconds for the object to fall to the ground. In conclusion, the highest altitude attained by the object is 576 feet, and it takes approximately 2.83 seconds for the object to fall to the ground.

Key concepts

Initial Velocity

When we examine projectile motion, one key concept to understand is 'initial velocity'. This is the speed at which an object is launched into the air. For our exercise, the initial velocity is specified as 128 ft/sec directed upward.

Understanding initial velocity is crucial because it influences two important aspects of projectile motion: the 'maximum height' an object reaches and the 'time of flight' or how long it stays in the air. The initial velocity directly affects the kinetic energy of the projectile. In the absence of air resistance, the only force acting on the object is gravity, which will decelerate the object at a constant rate until it reaches its peak height where the vertical velocity becomes zero.

Maximum Height

Moving on to the 'maximum height', which is the zenith of an object's journey upwards. In our exercise, we solved for the time when the vertical velocity is zero and then plugged that time into the position equation. The resulting value of 576 feet is the highest altitude the object reaches.

To visualize it, imagine the object climbing against gravity with its initial kinetic energy. As it ascends, gravity pulls it down, reducing its velocity. When all the initial kinetic energy is converted to potential energy, the object stops momentarily at the maximum height before descending back down.

Time of Flight

The 'time of flight' is another key term, and it refers to the duration the object remains airborne. In our exercise, we determined the time of flight by solving for when the object's height would return to zero using the position equation. Our solutions indicated two times, but the second time, 2.83 seconds, was the true 'time of flight' for the object to hit the ground after it was projected upward.

An understanding of this concept helps students predict how long an object will stay in the air, which is useful in various practical applications, such as in sports or aeronautics. The initial velocity and the force of gravity determine this flight duration.

Quadratic Formula

The 'quadratic formula' is a vital tool used to find the solutions of quadratic equations, which appear frequently in projectile motion problems. In our exercise, we used it to find the times at which the object reaches the ground level.

The formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides the solutions for the unknown value 't' in our equation, representing the time when the object's height is zero. The +/- sign indicates that there can be two possible times which correspond to two different physical events—the object reaching maximum height and the object returning to the ground. By understanding the quadratic formula, students can tackle a myriad of physics problems involving acceleration, motion, and forces.