2. A bullet was shot by a gun with an initial speed of 260 m/s and is fired at an angle of 30 to the ground. a. How much is the value Vy at the top of the bullet's trajectory? b. Calculate the time taken by the bullet to reach its maximum altitude. c. Deduce the total time taken by the bullet in the air. d. Calculate the range of the bullet.

Explanation:

a )

the value Vy at the top of the bullet's trajectory will be zero . At the top , only the horizontal component Vx will exist .

b )

At the top Vy = 0

initial velocity will have vertical component = 260 sin30 = 130 m /s

acceleration = - g = - 9.8

v = u - gt

o = 130 - 9.8 t

t = 13.26 s .

c )

Total time = Time of ascent + time of descent

= 13.26 + 13.26 = 26.52 s

d )

Range R = u² sin2Ф / g , where u is initial velocity , Ф is angle of throw , g is acceleration due to gravity .

R = 260² sin 2 x 30 / 9.8

= 5973.8 m .

f=ma newton's second law 7x8=56n

the maxium height is 34m/s