Question

A player can throw a ball to a maximum horizontal distance of 80 m. If he throws the ball vertically with the same velocity, then the maximum height reached by the ball is: 

• A. 160
• B. 60
• C. 20
• D. 40

Answer 1

The Correct Option is D

To solve the problem, we need to find the maximum height reached by a ball thrown vertically upward with the same velocity that gives a maximum horizontal range of 80 m when thrown at an optimal angle.

1. Understanding the Projectile Range Formula:
The maximum horizontal range of a projectile (when projected at an angle of 45°) is given by:
$ R = \frac{u^2}{g} $
where:
$R$ = range (80 m),
$u$ = initial velocity,
$g$ = acceleration due to gravity ($9.8 \, \text{m/s}^2$).

2. Rearranging the Formula to Find $u$:
We rearrange the equation to solve for $u^2$:
$ u^2 = R \cdot g = 80 \cdot 9.8 = 784 $

3. Using Vertical Motion Formula to Find Maximum Height:
When the ball is thrown vertically with the same velocity, the maximum height $H$ is given by:
$ H = \frac{u^2}{2g} $
Substitute the value of $u^2$:
$ H = \frac{784}{2 \cdot 9.8} = \frac{784}{19.6} = 40 \, \text{m} $

Final Answer:
The maximum height reached by the ball is 40 meters.

Answer 2

The correct option is: (D) 40.

When a projectile is launched with the same initial velocity in both the horizontal and vertical directions, the time of flight for both motions is the same. In this scenario, the horizontal distance traveled by the projectile is the same in both cases.

Given that the player can throw the ball to a maximum horizontal distance of 80 meters, this means that the time of flight for the vertical motion (maximum height) will be the same.

Since the vertical motion is affected by gravity, the time taken to reach the maximum height will be half of the total time of flight. Therefore, if the time of flight is the same for both motions, the time taken to reach the maximum height is half of the total time.

Thus, the maximum height reached by the ball will be half of the time of flight times the vertical component of the initial velocity. In this case, the maximum height would be:

Maximum Height = (1/2) * Time of Flight * Initial Vertical Velocity

Since the horizontal distance is 80 meters and we know that horizontal distance (d) is given by:

d = Horizontal Velocity * Time of Flight

We can rearrange this to solve for Time of Flight:

Time of Flight = d / Horizontal Velocity

Plugging in the values:

Time of Flight = 80 m / Horizontal Velocity

Now, since the initial vertical velocity is the same as the initial horizontal velocity:

Maximum Height = (1/2) * (80 m / Horizontal Velocity) * Initial Vertical Velocity

This simplifies to:

Maximum Height = (1/2) * 1 * Initial Vertical Velocity = Initial Vertical Velocity / 2

So, the maximum height reached by the ball is indeed half of the initial vertical velocity, which means it's 40 meters, as given in the answer.