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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?
Physics
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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We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

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