Question

Guys please help me ​

Answer 1

1)
`t=2.26\: s`

2)
`S=33.9\: m`

3)
`v=26.77\: m/s`

4)
`\alpha=55.92`

Step-by-step explanation:

1)

We can use the following equation:

`y_(f)=y_(0)+v_(iy)t-0.5*g*t^(2)`

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

`0=25-0.5*9.81*t^(2)`

`t=2.26\: s`

2)

The equation of the motion in the x-direction is:

`v_(ix)=(S)/(t)`

`15=(S)/(2.26)`

`S=33.9\: m`

3)

The velocity in the y-direction of the stone will be:

`v_(fy)=v_(iy)-gt`

`v_(fy)=0-(9.81*2.26)`

`v_(fy)=-22.17\: m/s`

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

`v=\sqrt{v_(x)^(2)+v_(fy)^(2)}=\sqrt{15^(2)+(-22.17)^(2)}`

`v=26.77\: m/s`

4)

The angle of this velocity is:

`tan(\alpha)=(22.17)/(15)`

`\alpha=tan^(-1)((22.17)/(15))`

`\alpha=55.92`

Then α=55.92° negative from the x-direction.

I hope it helps you!