Question

A ball is thrown vertically upward with a speed of 49 m/s. How long will it take to return to the thrower’s hand?

• A. 5 s
• B. 7 s
• C. 10 s
• D. 14 s

Hint:

Use the formula \( t = \frac{u}{g} \) for vertical motion to calculate time to peak, and double it for total time.

Answer 1

The Correct Option is C

Step 1: Identify given values
- Initial velocity, \( u = 49 \, \text{m/s} \) (upward)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \) (downward, so we take \( g \) as positive in the downward direction)
- Final velocity at the highest point, \( v = 0 \, \text{m/s} \)

Step 2: Calculate time to reach the highest point
Using the kinematic equation:
\[ v = u - gt \] Since the ball is moving upward, we take \( u \) as positive and \( g \) as positive (downward acceleration):
\[ 0 = 49 - 9.8 \cdot t \] \[ t = \frac{49}{9.8} = 5 \, \text{seconds} \] So, it takes 5 seconds to reach the highest point.

Step 3: Calculate total time
The time to go up equals the time to come down, so the total time is:
\[ \text{Total time} = 2 \times 5 = 10 \, \text{seconds} \]