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

Answer:

x = 15√3m , y = 3.75m

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 .