Question

11. A shell of mass 5 kg is fired at 60° to the horizontal with a speed of 20 m s. Neglecting the air resistance, the K. E of shell in joules at its highest point: a. Zero b. 25 1/6 d. 250 18 C é -
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Answer 1

Answer:

d)   250 J

Explanation:

K.E = $1/2 mv^2$                                                                                                                              K.E = $1/2 mivi^2$Cos θ

in this case we are given he angle theta with horizontal axis 1st we will convert it into angle wth y axis by simply subtrcting 60⁰ from90⁰ that become 30° . NOW putting in equation.  

K.E= $1/2m(v Cos 30)$ ²

K.E =1/2 mv² (1/2)²

K.E = KE (1/4)

frm the statement the KE is 1000 J.

By dividing it with 4 we get 250 J.