Question

1.) A ball is thrown horizontally from the top of a building. What is the vertical velocity of the ball after 1 second?

Answer 1

Answer:

1. B. 9.8 m/s
2. A. 0 m/s²
3. C. 45°
4. B. 70°
5. D. uniformly accelerated motion

Explanation:

1.

u=0m/s

v=?

a=9.8m/s²

t=1sec

now

we have

v=u+at

v=0+9.8*1

v=9.8m/s

3.

angle is measured in sin and cos angle where maximum angle's value is gained by 45°

4.

we have

range

R=$\frac{u²sin2\theta}{g}$

For

angle =20°

R1=$\frac{u²sin2*20}{g}$.......[1]

another angle be $\theta$

R2=$\frac{u²sin2\theta}{g}$.......[2]

since R1=R2

By comparing we get

Sin40=Sin($2\theta$)

Sin(180-40)=Sin($2\theta$)

we

140=$2\theta$

$\theta$=70°

Answer 2

1. 9.8
2. zero
3. 45
4. not sure - 70
5. Uniform accelerated motion