Question

a stone is projected with a velocity of 20 metre per second at an angle of 30° horizontal after 1.5 seconds find its horizontal distance and vertical height for it starting point​

Answer 1

x = 15√3m , y = 3.75m

Step-by-step explanation:

According to the question ,

• Initial velocity = 20m/s
• Angle of projection = 30°

And we need to find out ,

• Its horizontal distance and vertical height for it starting point after 1.5s .

As we know that , at any time t , the horizontal displacement (x) is given by ,

`\longrightarrow` x = u cos
`\theta`t

`\longrightarrow` x = 20 * cos30° * 1.5 m

`\longrightarrow`x = 20 * √3/2 * 3/2 m

`\longrightarrow` x = 15√3 m .

Hence the horizontal distance is 15√3m from the starting point after 1.5s .

`\rule{200}2`

For finding the vertical distance , let's find range first .

`\longrightarrow` R = u²sin2
`\theta`/g

`\longrightarrow` R = (20)² * sin60° / 10 m

`\longrightarrow`R = 400 * √3/10 *2

`\longrightarrow`R = 20√3 m

Now we can find the vertical displacement as ,

`\longrightarrow` y = xtan
`\theta` ( 1 - x/R)

`\longrightarrow`y = 15√3 *1/√3 ( 1-15√3/20√3)m

`\longrightarrow`y = 15( 1 - 3/4)m

`\longrightarrow`y = 15 * 1/4m

`\longrightarrow` y = 3.75 m

Hence the vertical distance is 3.75m from the ground after 1.5s .