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A projectile is thrown with a velocity v...

A projectile is thrown with a velocity `vec v _(0) = 3hati + 4 hatj ms ^(-1)` where `hat I and hatj` are the unit vectors along the horizontal and vertical directions respectively. The speed of the projectile at the highest point of its motion in `ms ^(-1)` is

A

3

B

4

C

5

D

`0.75`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the motion of the projectile and determine its speed at the highest point of its trajectory. ### Step-by-Step Solution: 1. **Identify the Initial Velocity Components:** The initial velocity of the projectile is given as \( \vec{v_0} = 3 \hat{i} + 4 \hat{j} \) m/s. Here, the horizontal component \( v_{0x} = 3 \) m/s and the vertical component \( v_{0y} = 4 \) m/s. 2. **Understand Projectile Motion:** In projectile motion, the horizontal component of velocity remains constant throughout the motion, while the vertical component changes due to the acceleration due to gravity. 3. **Determine the Velocity at the Highest Point:** At the highest point of the projectile's motion, the vertical component of the velocity becomes zero (\( v_y = 0 \)) because the projectile stops rising and is about to start falling down. 4. **Calculate the Speed at the Highest Point:** Since the vertical component of velocity is zero at the highest point, the speed of the projectile at this point is equal to the horizontal component of the velocity: \[ v = v_{x} = 3 \text{ m/s} \] 5. **Final Answer:** Therefore, the speed of the projectile at the highest point of its motion is \( 3 \) m/s.
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