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SVETLANKA909090 [29]

3 years ago

6

A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of

the trajectory? (b) What is its time of flight?

1 answer:

Fudgin [204]3 years ago

7 0

Answer:

Part a)

$v_f = v_x = 32.77 m/s$

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

$v_x = 40 cos35 = 32.77 m/s$

$v_y = 40 sin35 = 22.94 m/s$

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

$v_f = 32.77 m/s$

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

$\Delta y = 0$

$0 = v_y t - \frac{1}{2}gt^2$

$T = \frac{2v_y}{g}$

$T = \frac{2(22.94)}{9.81} = 4.68 s$

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According to the <u>Third Kepler’s Law</u> of Planetary motion:

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$a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}$ (3)

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3 years ago

A 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

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Their velocity afterward is v=3.467 m/s

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Final velocity of both players=?

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mathematically represented as

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ...........................(1)

where

$u_1$ =intial velocity of the football player

$u_2$ = inital velocity of second football player

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$v_2$ =final velocity of second football player

after tackling , both the football players moves with the same velocity,

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$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{(m_1+m_2)}$

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