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Difficulty: MediumKinematics and Linear Motion

A ball is thrown vertically upwards with an initial velocity of 30 m/s30\text{ m/s} from the edge of a cliff that is 35 m35\text{ m} above ground level. Taking the acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2, what is the total time elapsed before the ball hits the ground?

  1. A
    3.0 s3.0\text{ s}
  2. B
    4.0 s4.0\text{ s}
  3. 7.0 s7.0\text{ s}Answer
  4. D
    5.0 s5.0\text{ s}

Answer

The total time elapsed before the ball hits the ground is 7.0 s7.0\text{ s}.
Using the displacement equation s=ut−12gt2s = ut - \frac{1}{2}gt^2 with downward taken as negative, the net displacement when the ball reaches the ground is −35 m-35\text{ m}. Setting up the equation: −35=30t−5t2-35 = 30t - 5t^2, which simplifies to t2−6t−7=0t^2 - 6t - 7 = 0. Factoring gives (t−7)(t+1)=0(t - 7)(t + 1) = 0, yielding t=7.0 st = 7.0\text{ s}.

Step-by-Step Solution

1
Calculate the time taken (t1t_1) to reach maximum height
t1=ug=3010=3.0 st_1 = \frac{u}{g} = \frac{30}{10} = 3.0\text{ s}
At maximum height, the final vertical velocity is 0 m/s0\text{ m/s}.
2
Calculate the maximum height (hmaxh_{max}) reached above the cliff
hmax=u22g=3022(10)=45 mh_{max} = \frac{u^2}{2g} = \frac{30^2}{2(10)} = 45\text{ m}
Using the kinematic equation v2=u2−2ghv^2 = u^2 - 2gh.
3
Determine total height above ground and time (t2t_2) to fall to the ground
Total height H=35 m+45 m=80 mH = 35\text{ m} + 45\text{ m} = 80\text{ m}. t2=2Hg=2(80)10=4.0 st_2 = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2(80)}{10}} = 4.0\text{ s}
The ball falls from rest from a peak height of 80 m80\text{ m}.
4
Calculate the total time of flight
ttotal=t1+t2=3.0 s+4.0 s=7.0 st_{total} = t_1 + t_2 = 3.0\text{ s} + 4.0\text{ s} = 7.0\text{ s}
The complete journey consists of ascending to the peak and descending to the ground.

Key Concept

Kinematics of Vertical Motion Under Gravity
Estimated Time:1m 30s
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Topics

Kinematics and Linear Motion

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