A ball is launched from the top of a building at an initialspeed of 45 m\\sec an

Question

A ball is launched from the top of a building at an initialspeed of 45 msec and at an angle of 37 degrees above thehorizontal axis. The building is 25.5 meters above the horizontallevel. Assume the air effects can be ignored. a- calculate the horizontal and vertical components of theprojectiles initial velocity b- calculate the time of flight of the projectile in theair c- what are the horizontal and vertical components of theprojectiles velocity as it strikes the ground d- What is the magnitude of the projectiles velocity as itstrikes the ground e- what is the angle between the projectiles velocityand horizontal direction at the instant it strikes the ground f- horizontal distance of x A ball is launched from the top of a building at an initialspeed of 45 msec and at an angle of 37 degrees above thehorizontal axis. The building is 25.5 meters above the horizontallevel. Assume the air effects can be ignored. a- calculate the horizontal and vertical components of theprojectiles initial velocity b- calculate the time of flight of the projectile in theair c- what are the horizontal and vertical components of theprojectiles velocity as it strikes the ground d- What is the magnitude of the projectiles velocity as itstrikes the ground e- what is the angle between the projectiles velocityand horizontal direction at the instant it strikes the ground f- horizontal distance of x

Explanation / Answer

the horizontal initial velocity=45*cos37=35.94 the vertical initial velocity=45*sin37=27.08 vertical accleration=g=9.8 horizontal accleration=0 the vertical velocity becomes 0 at the max. height let projectile reached the max. height after t sec after it waslaunched v=u-g*t 0=27.08-9.8*t t=2.76 sec v2=u2-2*g*h 0=27.082-2*9.8*h h=37.41 m let the projectile took T sec to reach the ground from the max.height h=u*t+(1/2)*g*T2 37.41+25.5=0+(1/2)*9.8*T2 T=3.58 total time of flight=t+T=2.76+3.58=6.34 sec (ans b) let vertical velocity as it hit the ground=v1 v1=u+g*T v1=0+9.8*3.58 v1=35.084 m/s and since the horizontal velocity does not under goes andaccleration it remains same and is equal to 35.94 magnitude of the projectile as it strike theground=(35.942+35.0842)=50.23 m/s(ans) let the projectile velocity make an angle with thehorizontal as it strike the ground horizontal velocity as the projectile strike theground=35.94 50.23 cos=35.94 =44.32 (ans e) horizontal distance=35.94*6.34=227.86 m (ans)

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