1 An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is thrown? Take g=10ms−2

15 m above the top of the building
30 m below the top of the building
15 m below the top of the building
30 m above the building

2 A stone thrown from ground level returns to the same level 4 s after. With what speed was the stone thrown? Take g=10ms−2
20 m/s
10 m/s
30 m/s
15 m/s

3 What is common to the variation in the range and the height of a projectile?
horizontal velocity
time of flight vertical velocity, horizontal acceleration
vertical velocity
horizontal acceleration

4 A cart is moving horizontally along a straight line with constant speed of 30 m/s. A projectile is fired from the moving cart in such a way that it will return to the cart after the cart has moved 80 m. At what speed (relative to the cart) and at what angle (to the horizontal) must the projectile be fired?
35.8 m/s at 24 degrees
38.6 m/s at 54 degrees
27 m/s at 35 degrees
24 m/s at 44 degrees

5 The trajectory of a projectile is
an ellipse
a circle
a parabola
a straight line

6 The motion of a ball rolling down a ramp is one with
constant speed
increasing acceleration
constant acceleration
decreasing acceleration

7 How fast must a ball be rolled along the surface of a 70-cm high table so that when it rolls off the edge it will strike the floor at the same distance (70cm) from the point directly below the edge of the table?
174.5 cm/s
185.2 cm/s
215.3 cm/s
143.7 cm/s

8 A ball is kicked and flies from point P to Q following a parabolic path in which the highest point reached is T. The acceleration of the ball is
zero at T
greatest at P
greatest at T and Q
the same at P as at Q and T

9 A mass accelerates uniformly when the resultant force acting on it
is zero
is constant but not zero
increases uniformly with respect to time
is proportional to the displacement of the mass from a fixed point

10 The term that best describes the need to hold the butt of a riffle firmly against the shoulder when firing to minimise impact on the shoulder is
forward displacement
forward acceleration
recoil velocity
pressure

11 Two trolleys X and Y with momenta 20 Ns and 12 Ns respectively travel along a straight line in opposite directions before collision. After collision the directions of motion of both trolleys are reversed and the magnitude of the momentum of X is 2 Ns. What is the magnitude of the corresponding momentum of Y?
6 Ns
8 Ns
10 Ns
30 Ns

12 A force of 2i+7j N acts on a body of mass 5kg for 10 seconds. The body was initially moving with constant velocity of i−2j m/s. Find the final velocity of the body in m/s, in vector form.
5i+12j
12i−5j
10i−7j
7i+10j

13 The exhaust gas of a rocket is expelled at the rate of 1300 kg/s, at the velocity of 50 000 m/s. Find the thrust on the rocket in newtons
65107
35107
76107
57107

14 Sand drops at the rate of 2000 kg/min. from the bottom of a hopper onto a belt conveyor moving horizontally at 250 m/min. Determine the force needed to drive the conveyor, neglecting friction.
500 N
800 N
139 N
152 N

15 A 30,000-kg truck travelling at 10.0m/s collides with a 1700-kg car travelling at 25m/s in the opposite direction. If they stick together after the collision, how fast and in what direction will they be moving?
8.1 m/s in the direction of the truck's motion
12.3 m/s in the direction of the car's motion
24.2 m/s in the direction of the car's motion
17.6 m/s in the direction of the truck's motion

16 A gun of mass M is used to fire a bullet of mass m. The exit velocity of the bullet is v. Find the recoil velocity of the gun
Mv/m
mv/M
–Mvm
−mvM

17 A 40-g ball travelling to the right at 30 cm/s collides head on with an 80-g ball that is at rest. If the collision is perfectly elastic, find the velocity of each ball after collision
the first ball is going to the right at 10m/s while the other is going to the left at 20m/s
the first ball is going to the left at 10m/s while the other is going to the right at 20m/s
the first ball is going to the left at 20 m/s while the other is going to the right at 10 m/s
the first ball is going to the right at 10 m/s while the other is going to the left at 10 m/s

18 A 10-g bullet of unknown speed is shot horizontally into a 2-kg block of wood suspended from the ceiling by a cord. The bullet hits the block and becomes lodged in it. After the collision, the block and the bullet swing to a height 30cm above the original position. What was the speed of the bullet? (This device is called the ballistic pendulum). Take g=98ms−2
487 m/s
640 m/s
354 m/s
700 m/s

19 How large an average force is required to stop a 1400-kg car in 5.0 s if the car’s initial speed is 25 m/s?
2000 N
3500 N
9000 N
7 000N

20 Which of these is NOT a statement of Newton’s law of universal gravitation?
gravitational force between two particles is attractive as well as repulsive
gravitational force acts along the line joining the two particles
gravitational force is directly proportional to the product of the masses of the particles
gravitational force is inversely proportional to the square of the distance of the particles apart

1. V = Vo + g*t.

Tr = (V-Vo)/g = (0-10)/-10 = 1 s. = Rise
time or time to reach max. ht.

Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg.

3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg.

d = Vo*t + 0.5g*t^2.
d = 10*1 + 5*1^2 = 15 m. Below top of bldg.

2. Tr = 4/2 = 2 s. = Rise time.

V = Vo + g*t.
Vo = V - g*t = 0 - (-10)*2 = 20 m/s.

3. Vertical Velocity.

4. V*t = 80 m.
30t = 80
T = 2.67 s. = Time in flight.

Tr = T/2 = 2.66/2 = 1.33 s. = Rise time.

Y = Yo + g*t.
Yo = Y-g*t = 0-(-10)*1.33 = 13.3 m/s. =
Ver. component of initial velocity.

tanA = 13.3/30 = 0.443333
A = 24o.

Vo = 30/cos24 = 32.8 m/s @ 24o.

5. Parabola.

6. Constant acceleration.

7. Vo*t + 0.5g*t^2 = 0.7 m.
4.9t^2 = 0.7
t^2 = 0.1429
29Tf = 0.378 s = Fall time or time to hit the floor.

Vx * 0.378 = 0.7 m.
Vx = 1.852 m/s. = 185.2 cm/s.

yes

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t=3s g=10m/s2 v=10m/s

s=ut-1/2gt2
s=10*3-1/2*10*3^2
s=30-45
s=-15m ....(-) means below the top down
:s=15m

pls i need answer no 1

An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is thrown? Take g=10ms−2.
pls answer the answer for me

plz ans the question no.11

1. To solve this problem, we can use the equation of motion for objects in free fall:

h = u*t + (1/2) * g * t^2

where:
h = height
u = initial velocity
t = time
g = acceleration due to gravity

In this case, the object is thrown upward, so the initial velocity is positive (10 m/s) and the acceleration due to gravity is negative (-10 m/s^2).

Plugging in the values, we get:
h = 10 * 3 + (1/2) * -10 * (3)^2
h = 30 - 45 = -15

Since the height is negative, it means the object is below the top of the building. Therefore, the correct answer is 15 m below the top of the building.

2. To solve this problem, we again use the equation of motion for objects in free fall. However, in this case, the object returns to the same level, so the height is zero.

h = u*t + (1/2) * g * t^2

Plugging in the values, we get:
0 = u * 4 + (1/2) * -10 * (4)^2
0 = 4u - 80
4u = 80
u = 20 m/s

Therefore, the stone was thrown with a speed of 20 m/s.

3. The parameter that is common to the variation in the range and the height of a projectile is the time of flight. The time of flight is the total time during which the projectile is in motion. Both the range and the height depend on the time of flight. Other parameters such as the initial velocity, angle of projection, and acceleration due to gravity can affect the range and height, but the time of flight is directly related to both.

4. To solve this problem, we need to use the concept of relative velocity.

We know that the cart moves with a constant speed of 30 m/s. Let's assume the projectile is fired with a velocity of v at an angle θ to the horizontal.

Since the projectile has to return to the cart after the cart has moved 80 m, the horizontal velocity of the projectile should be equal to the horizontal velocity of the cart. This means we have:

v * cos(θ) = 30 m/s

To find the vertical velocity, we can use the equation of motion for vertical motion:

s = u * t + (1/2) * g * t^2

where:
s = 80 m (horizontal distance the cart moves)
u = v * sin(θ) (initial vertical velocity)
t = time of flight

Since the projectile should return to the same height it was launched from, the equation becomes:

0 = u*t + (1/2) * g * t^2

Solving for t, we get:

t = 2u / g

Substituting the values:

80 = (v * sin(θ)) * (2 * v * sin(θ)) / g

80 = 2 * (v^2 * sin^2(θ)) / g

v^2 * sin^2(θ) = 40 * g

v * sin(θ) = √(40 * g)

Now we can plug this value into the equation for the horizontal velocity:

v * cos(θ) = 30 m/s

Substituting v * sin(θ) = √(40 * g), we get:

√(40 * g) * cos(θ) = 30 m/s

cos(θ) = 30 / √(40 * g)

θ = cos^(-1)(30 / √(40 * g))

Finally, we can calculate the values:

v = √(30^2 + (√(40 * g))^2)
θ = cos^(-1)(30 / √(40 * g))

After calculating, the values are:
v ≈ 38.6 m/s
θ ≈ 54 degrees

Therefore, the projectile must be fired with a speed of 38.6 m/s at an angle of 54 degrees to the horizontal.

5. The trajectory of a projectile is a parabola.