A man throws a ball withinitial Velocity=10m/sfrom a building with height h=600m

Question

A man throws a ball withinitial Velocity=10m/sfrom a building with height h=600m. A second man letsgo of a ball v=0m/s at a height of 575m.

Given acceleration ay= -g=-9.8m/s^2

mass of ball 1= 1.50kg

mass of ball 1= 5.00kg

What is the time that both balls hit the ground? asuming both times will be the same since regardless of mass and speed gravity most important .

What is the velocity of ball 1 before it hits ground? vfinal 1

What is the velocity of ball 1 before it hits ground? vfinal 2

Explanation / Answer

FOR BALL 1 s=ut+1/2at^2 600=10*t+(0.5*9.8)*t^2 Solving for t 10.09seconds Vfinal=u+at Vfinal= 10+(9.8*10.09) Vfinal=108.882 FOR BALL 2 s=1/2at^2 => 575=(0.5*9.8)*t^2 Solving for t => t=10.83seconds Vfinal=u+at Vfinal=0+(9.8*10.83) Vfinal=106.134 TIME FOR BALL 1=10.09sec TIME FOR BALL 2=10.83sec FINAL SPEED OF BALL 1=108.882 m/sec FINAL SPEED OF BALL 2=106.134 m/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.