Question

a body is projected with a velocity of 30m/s at an angle 60 degree .find the maximum height ,time horizontal range ,full answer with steps

Answer 1

ANSWER :-

Maximum height,

H = u² sin²β/2g = 30²× sin²30°/(2×10) = 11.25 m

Time of flight ,

T = 2u sinβ /g = 2×30 ×sin30°/10 =3 sec

Horizontal range,

R= u²sin2β/g = 30² × sin60°/10 = 45√3 m

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