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2 years ago

13

A ball is thrown with an initial speed of 10. Meters per second. At what angle above the horizontal should the ball be thrown to

reach the greatest height?

2 answers:

maria [59]2 years ago

6 0

The initial speed doesn’t matter. To reach the maximum height, you need to launch straight up ... 90 degrees above horizontal.

Jlenok [28]2 years ago

5 0

Answer:

90 degree

Explanation:

The height for the projectile motion is given by

H = u^2 Sin^2 theta/ 2g

Height is maximum when sin theta is maximum.

Sin theta is maximum when it's value is 1.

So the value of theta is 90 degree.

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Read 2 more answers

With an initial velocity of 20km per hour a car accelerated at 8m/s2 for 10s,what is the position of the car at the end of 10s

Dima020 [189]

initial velocity=u=20km/h=5.5m/s
Acceleration=a=8m/s^2
Time=t=10s

$\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf\longmapsto s=5.5(10)+\dfrac{1}{2}(8)(5.5)^2$

$\\ \sf\longmapsto s=55+4(5.5)^2$

$\\ \sf\longmapsto s=55+4(30.25)$

$\\ \sf\longmapsto s=55+121$

$\\ \sf\longmapsto s=176m$

4 0

2 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago