Question

a ball is thrown vertically down from the edge of a cliff with a speed of 5.2m/s how high is the cliff if it took 11s for the ball to reach the ground​

Answer 1

• initial velocity=u=5.2m/s
• Final velocity=v=0m/s
• Time=t=11s
• Height=h
• Acceleration=g=9.8m/s^2

`\\ \sf\longmapsto h=ut+(1)/(2)gt^2`

`\\ \sf\longmapsto h=5.2(11)+(1)/(2)(9.8)(11)^2`

`\\ \sf\longmapsto h=57.2+4.9(121)`

`\\ \sf\longmapsto h=57.2+592.9`

`\\ \sf\longmapsto h=650.1m`