An object is thrown directly downwards from a height of 60 m with an initial velocity of 11 m/s. What will be its velocity on impact?

An object is thrown directly downwards from a height of 60 m with an initial velocity of 11 m/s. What will be its velocity on impact?

• A. 34.5 m/s
• B. 1,289.16 m/s
• C. 24.2 m/s
• D. 36.03 m/s

Answer: (D)

To determine the object's velocity upon impact when thrown directly downwards from a height of 60 m with an initial velocity of 11 m/s, we can employ the kinematic equations of motion. The equation most relevant in this case is: \[ v^2 = u^2 + 2gh \] Where: - \( v \) is the final velocity upon impact. - \( u \) is the initial velocity (11 m/s). - \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). - \( h \) is the height from which the object is thrown (60 m). First, we convert the height into the equation with gravity acting downwards (positive), leading to: \[ v^2 = (11 \, \text{m/s})^2 + 2 \cdot 9.81 \, \text{m/s}^2 \cdot 60 \, \text{m} \] Calculating the initial velocity term: \[ (11 \, \text{m/s})^2 = 121 \, \text{m}^2/\text{s}^2 \] Calculating the gravitational term: \[ 2 \

To determine the object's velocity upon impact when thrown directly downwards from a height of 60 m with an initial velocity of 11 m/s, we can employ the kinematic equations of motion. The equation most relevant in this case is:

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

Where:

• ( v ) is the final velocity upon impact.

• ( u ) is the initial velocity (11 m/s).

• ( g ) is the acceleration due to gravity (approximately 9.81 m/s²).

• ( h ) is the height from which the object is thrown (60 m).

First, we convert the height into the equation with gravity acting downwards (positive), leading to:

[ v^2 = (11 , \text{m/s})^2 + 2 \cdot 9.81 , \text{m/s}^2 \cdot 60 , \text{m} ]

Calculating the initial velocity term:

[ (11 , \text{m/s})^2 = 121 , \text{m}^2/\text{s}^2 ]

Calculating the gravitational term:

[ 2 \