Question

A projectile is fired with an initial speed of 20. m/s at an angle of 42.2° above the horizontal on a long flat firing range. Determine,
a) the maximum height reached by the projectile,
b) the total time in the air,

c) the total horizontal distance covered (that is, the range),

d) the speed of the projectile 1.50 s after firing.
e) direction of the speed at 1.50 s.

Answer 1

Anos :- n Ny=0 ry usino u - - - - - Ottomat uz Initial ucose In a direction х = cogee Teese 49 тоо acceleration in x-direction - In y direction Fayation of motion between initial At Final Poortyon A Poristive - s = ut t I at => 0= scos usino ttl (8) I 22 Negative se : f= zusino an - Time of feight Total horizontal distance It covered (R) = ucos out = 4800g 0 x 24 sino - Range g = uz sunce con =) y = Yytat i => y = usino-get Uy² = y² +299 o= 42sin²o+26-gH H = y2sino < 29 maximum Height

Now (9) (6) H= y²sinzo = (20) sim2(42.2) 29 2x9.8 9.208 mles T= 2usino = 2x20 sin(4202) g gog = 20742 sec R = yz sin (20) . (c) = (20)2 Sin (2x42 2) - 40.62 m (d) x = 4x = 4coso = zocos (42.2) = 19.985 miss y = 4y - get = 4syno -9.8 x1.5 = 2osin (42.2) - 9.881-5= -1.265 m/s speed at los sec = I x² + y² = J (19.985)2 + (-10265)2 200025 mles (e) For direction x tana= uy = 1.265 vy Vx 19.985 2= 3.62 whith Possitive x-axis in clockwise dyrection .