Question

A ball thrown straight up takes \(2.25 \mathrm{~s}\) to reach a height of \(36.8 \mathrm{~m} .\) (a) What was its initial speed? ( \(b\) ) What is its speed at this height? ( \(c\) ) How much higher will the ball go?

Short answer

The initial speed of the ball is approximately 26.3 m/s. The ball is still traveling upwards with a speed of approximately 4.3 m/s at a height of 36.8m. The ball will ascend an additional approximate 0.94m, resulting in a total approximate height of 37.7m.

Step-by-step solution

Compute the Initial Speed

Rearranging the height formula to solve for the initial speed 'u', we have \(u = (H - 0.5at^2) / t\). Given \(H = 36.8m\), \(a = - 9.8 m/s^2\) due to gravity, and \(t = 2.25\), solve for \(u\).

Calculate the speed at this height

To calculate the speed when the ball is at the height of 36.8m, substitute \(u\), \(a\), and \(t\) into the velocity equation \(V = u + at\). Here the state of the ball is still going up.

Estimate the Maximum Height

When the ball reaches its maximum height, its speed is 0. Substituting \(V = 0\), \(a = - 9.8m/s^2\), and \(u\) into the velocity equation, we can derive the total time 'T' that the ball takes to reach the peak. After that we substitute \(T\), \(a\), and \(u\) into the height formula to find the maximum height. Substracting 36.8 from the maximum height will give us the additional height.

Key concepts

Initial Speed Calculation

Understanding how to compute the initial speed of a projectile is fundamental in physics. Let's break it down using the context of the exercise where a ball is thrown straight up and takes 2.25 seconds to reach a height of 36.8 meters.

To find the initial speed, we need to reorganize the kinematic equation for vertical motion: \[ u = \frac{H - 0.5at^2}{t} \] where

• \( u \) is the initial speed,
• \( H \) is the height reached,
• \( a \) is the acceleration (in this case, the acceleration due to gravity, which is \( -9.8 \text{m/s}^2 \) ),
• \( t \) is the time taken to reach that height.

In our problem, we already have \( H = 36.8 \text{m} \) and \( t = 2.25 \text{s} \). Plugging these values into the formula calculates the initial speed required for the ball to reach 36.8 meters after 2.25 seconds.

Vertical Velocity

The vertical velocity of a projectile is the speed at which it moves along the vertical axis of its trajectory. In the provided example, where the ball has been thrown upwards, we can quantify this at any point using the simple equation: \[ V = u + at \]

The variables are defined as follows:

• \( V \) is the velocity at a given time,
• \( u \) is the initial velocity,
• \( a \) is the acceleration, and
• \( t \) is the time elapsed since the beginning of the motion.

By substituting the previously calculated \( u \), along with \( a = -9.8 \text{m/s}^2 \) and \( t = 2.25 \text{s} \) into this equation, we find the speed of the ball at the 36.8-meter height. This velocity would be the speed at which the ball is still ascending before it starts to descend back to the ground.

Maximum Height Projectile

The maximum height of a projectile is a point where its vertical velocity becomes zero; it is the peak of its trajectory. To calculate this pinnacle of a projectile's path, we apply the same vertical velocity formula, but we set \( V = 0 \) as the final velocity at the maximum height.

By rearranging the vertical velocity equation \( 0 = u + at \), we can solve for the time \( T \) it takes to reach this point. From there, the maximum height \( H_{max} \) can be computed using the height formula: \[ H_{max} = uT + 0.5aT^2 \]

In the given problem, once \( T \) is found using the initial velocity \( u \) and \( a = -9.8 \text{m/s}^2 \), this information is substituted back into the height formula to find \( H_{max} \). To determine how much higher the ball will go beyond the 36.8-meter mark, the difference between \( H_{max} \) and 36.8 meters is calculated.