(a)Find the time taken for a ball to land 100 m from the base of a cliff when it was shot off at 20 m/s (b) Find the height  (c) Find the time taken to land if velocity was doubled

Question 10 options:

5s ; 122.5 ;5s ;

5s ; 122.5m ;10s ;

5s ; 250m ;5s ;

10s ; 122.5 ;5s ;

Let's solve this step-by-step.

### (a) Find the time taken for a ball to land 100 m from the base of a cliff when shot off at 20 m/s.

The ball is projected horizontally, meaning its initial vertical velocity \( v_{y0} \) is 0, and the horizontal velocity \( v_x \) is 20 m/s. The horizontal distance \( d_x \) is 100 m.

Using the horizontal motion equation:
\[
d_x = v_x \cdot t
\]
Substituting the known values:
\[
100 = 20 \cdot t
\]
Solving for \( t \):
\[
t = \frac{100}{20} = 5 \, \text{seconds}
\]

So, the time taken for the ball to land is **5 seconds**.

### (b) Find the height of the cliff.

The vertical motion is governed by the following equation (since the initial vertical velocity is 0):
\[
d_y = \frac{1}{2} g t^2
\]
Where:
- \( d_y \) is the vertical distance (height of the cliff),
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( t = 5 \, \text{seconds} \).

Substituting these values into the equation:
\[
d_y = \frac{1}{2} \times 9.8 \times (5)^2
\]
\[
d_y = \frac{1}{2} \times 9.8 \times 25 = 122.5 \, \text{m}
\]

So, the height of the cliff is **122.5 m**.

### (c) Find the time taken to land if the velocity was doubled.

If the horizontal velocity is doubled, the new velocity \( v_x' \) is:
\[
v_x' = 2 \times 20 = 40 \, \text{m/s}
\]
The horizontal distance remains the same (100 m), so using the horizontal motion equation again:
\[
100 = 40 \cdot t'
\]
Solving for \( t' \):
\[
t' = \frac{100}{40} = 2.5 \, \text{seconds}
\]

However, the time to land depends on the vertical motion, and the vertical motion is unaffected by the horizontal velocity. As we calculated earlier, the vertical time for the ball to hit the ground is **5 seconds**.

So, the time taken to land, even when the horizontal velocity is doubled, remains **5 seconds**.

### Final Answer:

The correct option is:

**5s ; 122.5m ; 5s**

---

Would you like any further details or clarifications? Here are 5 related questions to expand your understanding:

1. How does horizontal velocity affect the time of flight in projectile motion?
2. What would happen if the ball were projected at an angle instead of horizontally?
3. How would air resistance change the calculations for the time and height?
4. If the ball were shot off from a higher cliff, how would the time to land change?
5. How do you calculate the horizontal range of a projectile launched at an angle?

**Tip**: In projectile motion, the horizontal and vertical components of motion are independent of each other. Only gravity influences the vertical motion.

(a) Find the time taken for a ball to land 100 m from the base of a cliff when it was shot off at 20 m/s (b) Find the height (c) Find the time taken to land if velocity was doubled.

Solve a projectile motion problem where a ball is thrown off a cliff at 20 m/s. Learn how to calculate the time taken for the ball to land 100 m away, the height of the cliff, and the effect of doubling velocity. Ideal for students in Grades 9-12 studying kinematics and physics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation