Question

A body is projected vertically upward with an initial velocity of 40 m/s. Calculate the maximum height reached by the body. (Take \( g = 9.8 \, \text{m/s}^2 \))

• A. 80.4 m
• B. 160.8 m
• C. 100 m
• D. 120 m

Hint:

To calculate the maximum height of a body projected vertically, use the equation \( h = \frac{u^2}{2g} \), where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity. This equation assumes the final velocity at maximum height is zero.

Answer 1

The Correct Option is A

We are asked to find the maximum height reached by a body projected vertically upward with an initial velocity of \( u = 40 \, \text{m/s} \). We are given the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \). To solve this, we can use one of the standard equations of motion for vertical motion. The equation we will use is: \[ v^2 = u^2 - 2gh \] Where: - \( v \) is the final velocity (which is 0 at maximum height, since the body momentarily stops before falling back down), - \( u \) is the initial velocity (given as 40 m/s), - \( g \) is the acceleration due to gravity (given as 9.8 m/s\(^2\)), - \( h \) is the maximum height (the value we need to calculate). ### Step-by-step Solution: 1. Final velocity at maximum height: At the maximum height, the final velocity \( v = 0 \, \text{m/s} \), as the body momentarily comes to rest before reversing direction. 2. Substitute known values into the equation: Now, substitute \( v = 0 \), \( u = 40 \, \text{m/s} \), and \( g = 9.8 \, \text{m/s}^2 \) into the equation: \[ 0 = 40^2 - 2 \times 9.8 \times h \] This simplifies to: \[ 0 = 1600 - 19.6h \] 3. Solve for \( h \): Rearranging the equation to solve for \( h \): \[ 19.6h = 1600 \] \[ h = \frac{1600}{19.6} = 80.4 \, \text{m} \] Thus, the maximum height reached by the body is \( 80.4 \, \text{m} \).