Question

Styles Paragraph A bullet is fired directly upward and reaches a height of 2150 meters() Find the initial velocity of the bullet in meters per second (m/s). (2) How many seconds (s) does it take to reach its maximum height? (3) How many seconds (s) is the bullet in the air? Hint- acceleration of gravity is 9.8 m/SA2 売(Ctrl). 扇,季.wl

Answer 1

We know the reaction, h = ut + 1/2 gt^2 & v^2 = u^2 + 2gh v = u + x h = height: u = initial speed v = final speed g = gravitational acceleration t = time For upward directional motion g will be replaced by -g. therefore v^2 = u^2 - 2gh At maximum height, v = 0 u^2 = 2gh u = squareroot 2gh = squareroot 2 times 9/8 times 2150 m/sec. = 205.28 m/sec Initial velocity of bullet = 205.28 m/s For required time t, we can use this reaction, v = u - gt therefore t = u/g = 205.28/9.8 s (since v = 0) = 20.95 sec Time required to reach the height = 20.95 sec \r\n Total time of height in air = time for up direction motion + time for down direction motion = (20.95 + 20.95)[since time for upward & downward motion will be same] = 41.9 sec