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A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

3 years ago

Pls help it’s easy ik but I want to make sure of them both I don’t have much time

LuckyWell [14K]

Answer:

To the right

Weight

Explanation:

8 0

2 years ago

What’s the unit for velocity?

torisob [31]

Answer:

meter per second

Explanation:

It could be any other unit such as yard or feet, put it will be whatever measure per second or whatever time.

Examples

feet per second

miles per hour

4 0

4 years ago

How many centimeters are in 3.50 feet?

Kazeer [188]

106.68 centimetres are in 3.50 feet

4 0

3 years ago

A rock is dropped from a garage roof from rest. the roof is 6.0 m from the ground. determine the velocity of the rock as it hits

Dmitrij [34]

From rest, a rock is dropped from a garage roof. The roof is 6.0 meters above ground level. The rock will reach the earth at a speed of 10.849 meters per second.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity.

it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

Given data:

V(Final velocity)=? (m/sec)

h(height)= 6.0 m

u(Initial velocity)=0 m/sec

g(gravitational acceleration)=9.81 m/s²

Newton's third equation of motion:

$\rm v_y^2 = u_y^2+ 2gh \\\\\rm v_y^2 = 0+ 2gh\\\\\ v_y= \sqrt{2\times 9.81 \ (m/s^2)\times 6.0 (m)} \\\\ v_y=10.849 \ m/sec$

Hence, the velocity of the rock as it hits the ground will be 10.849 m/sec.

To learn more about the velocity refer to the link ;

brainly.com/question/862972

#SPJ1

7 0

2 years ago

PLEASE HELP! I'LL GIVE BRAINLEST​

PLEASE HELP! I'LL GIVE BRAINLEST