An object is dropped from a height of 55 m. What will its velocity be on impact?

An object is dropped from a height of 55 m. What will its velocity be on impact?

• A. 32.85 m/s
• B. 43.57 m/s
• C. 56.43 m/s
• D. 72.12 m/s

Answer: (A)

To determine the velocity of an object just before impact when it is dropped from a height, we can use the principle of conservation of energy or the kinematic equations of motion. In this situation, we can apply the following kinematic equation, which relates distance, initial velocity, final velocity, and acceleration: \[ v^2 = u^2 + 2as \] Where: - \( v \) is the final velocity, - \( u \) is the initial velocity, - \( a \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), - \( s \) is the height from which the object is dropped. Assuming the object is simply dropped from rest, the initial velocity \( u \) is \( 0 \). The height \( s = 55 \, \text{m} \), and the acceleration \( a \) is \( 9.81 \, \text{m/s}^2 \). Plugging these values into the equation gives: \[ v^2 = 0 + 2 \cdot 9.81 \cdot 55 \] Calculating this: \[ v^2 = 2

To determine the velocity of an object just before impact when it is dropped from a height, we can use the principle of conservation of energy or the kinematic equations of motion. In this situation, we can apply the following kinematic equation, which relates distance, initial velocity, final velocity, and acceleration:

[ v^2 = u^2 + 2as ]

Where:

• ( v ) is the final velocity,

• ( u ) is the initial velocity,

• ( a ) is the acceleration due to gravity (approximately ( 9.81 , \text{m/s}^2 )),

• ( s ) is the height from which the object is dropped.

Assuming the object is simply dropped from rest, the initial velocity ( u ) is ( 0 ). The height ( s = 55 , \text{m} ), and the acceleration ( a ) is ( 9.81 , \text{m/s}^2 ). Plugging these values into the equation gives:

[ v^2 = 0 + 2 \cdot 9.81 \cdot 55 ]

Calculating this:

[ v^2 = 2