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

x = 15√3m , y = 3.75m

Explanation:

According to the question ,

And we need to find out ,

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

x = u cost

x = 20 * cos30° * 1.5 m

x = 20 * √3/2 * 3/2 m

x = 15√3 m .

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

R = u²sin2/g

R = (20)² * sin60° / 10 m

R = 400 * √3/10 *2

R = 20√3 m

Now we can find the vertical displacement as ,

y = xtan ( 1 - x/R)

y = 15√3 *1/√3 ( 1-15√3/20√3)m

y = 15( 1 - 3/4)m

y = 15 * 1/4m

y = 3.75 m

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

i already answered it by 34/s