All categories

2 years ago

A ball is thrown upwards with an initial velocity of 25.o m/s, what is the velocity of the ball at 11.9 m from the ground?

Physics

1 answer:

mylen [45]2 years ago

5 0

Answer:

v2 = 19.79m/s

Explanation:

Okay, lets write down everything we have;

v1 = 25m/s [up]

v2 = ?

d = 11.9 [up]

a = 9.8 [down]

So the formula we will use is v2^2 = v1^2 + 2ad. We also have to make acceleration negative, so everything can have the same direction of up:

v2^2 = v1^2 + 2ad

v2^2 = (25)^2 + 2(-9.8)(11.9)

v2^2 = 625 - 233.24

v2^2 = 391.76

v2 = 19.79m/s

You might be interested in

A potter’s wheel moves from rest to an angular speed of 0.20 rev/s in 23.8 s. Assuming constant angular acceleration, what is it

ser-zykov [4K]

Answer:

$0.053 rad/s^2$

Explanation:

0.2 rev/s = 0.2 rev/s * 2π rad/rev = 0.4π rad/s

Since the angular acceleration is assumed to be constant, and the wheel's angular speed is increasing from rest (0 rad/s) to 0.4π rad/s within 23.8s. Then the angular acceleration must be

$\alpha = \Delta \omega / \Delta t = \frac{0.4 \pi - 0}{23.8} = 0.053 rad/s^2$

8 0

2 years ago

If a freely suspended vertical spring is pulled in downward direction and then released, which type of wave is produced in the s

larisa [96]

Answer:

longitudinal wave

Explanation:

it is perpendicular to the direction of the wave

3 0

2 years ago

Read 2 more answers

How much power is needed to move a 2000 kg mass from the bottom of the 150 m talk great pyramid in Egypt up a ramp if the ramp h

docker41 [41]

Answer:

How Heavy? More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).

Explanation:

4 0

2 years ago

The diagrams show objects' gravitational pull toward each

zaharov [31]

I don’t know sorry then ya soo

4 0

3 years ago

A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr

dmitriy555 [2]

Answer:

1.69515 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m$

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m

Time = Distance / Speed

$\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds$

The reaction time cannot be more than 1.69515 seconds

4 0

2 years ago