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inn [45]
3 years ago
10

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

the cliff 3.0 s later. a. How high is the cliff? b. What was the maximum height of the ball? c. What is the ball's impact speed?
Physics
1 answer:
Veseljchak [2.6K]3 years ago
3 0

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.

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