Question

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

A. 0 m/s
B. 9.8 m/s
C. 4.9 m/s
D. 19.6 m/s

2.) This is a required question
What is the horizontal acceleration of a projectile?

A. 0 m/s2
B. 8.9 m/s2
C. 4.9 m/s2
D. 9.8 m/s2

3.) If air resistance is neglected, at what angle a projectile must be projected to attain the maximum range?

A. 40
B. 75
C. 45
D. 90

4.) Neglecting air resistance, the range of a projectile projected at an angle of 20 degrees is the same as that of _______ if they were released with the same velocity.

A. 50
B. 70
C. 60
D. 80

5.) What do you call the motion of an object with constant acceleration?

A. circular motion
B. uniform motion
C. constant motion
D. uniformly accelerated motion


CAN SOMEONE PLEASE HELP ASAP. THANK U

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Answer 1

1. 9.8
2. zero
3. 45
4. 70
5. Uniform accelerated motion

Explanation:

Q.1 - V = u + at

u = 0, a = g = 9.8, t = 1

V = 0+9.8*1 = 9.8 m/s

Q 2- In horizontal position the velocity remains same hence a= dv/t, since change in velocity is zero the acceleration also equal to zero

Q.3- After neglecting the air resistance at 45° angle the maximum range is obtained.

let's prove, consider the range have maximum angle at θ, now range is given by

R = u²Sin2θ/g where sin2θ can we written as 2sinθ.cosθ.

Now the range value is directly proportional to the product of sin & cos of firing angle & this maximum value is obtained at 45°

Q.4 The range of two angle will be same if and only if they give 90° on addition.

20+x = 90,

x = 90-20 = 70°

Q.5 The motion with constant acceleration is known as uniformly accelerated motion.

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Answer 2

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°