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  • A skydiver drops 50 metres  without friction after a fall. When the parachute opens, it decelerates at <img alt="2 \; m/s^{2}" src="https://entrancecorner.on

A skydiver drops 50 metres  without friction after a fall. When the parachute opens, it decelerates at 2 \; m/s^{2}It reaches the ground at a speed of  3 \; m/s. How high was he released on bail?

Option: 1

293 m


Option: 2

111m


Option: 3

91 m

 


Option: 4

182 m


Answers (1)

best_answer

Initially, the parachutist falls by gravity:  

v^2=2ah

Or, v^2=2\times 9.8\times50\ =\ 980\ m^2s^{-2}

He reaches the ground with a speed =3\frac{m}{s}

a=-2\ ms^{-2}

Or, (3)^2= u^2-2 \times 2 \times h_{1}

Or, 9\ =\ 980 - 4 h_{1}

Or, h_1=\ \frac{971}{4}

Or, h_1=242.75\ metre

Therefore, 

Total height, 

  \\=50\ +\ 242.75\ =\ 292.75\\\ \approx293

Posted by

manish

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