Question

an airplane carrying out a parcel drops releases a parcel while travelling at a steady speed of 90 m/s at an altitude of 200 m. caculate a)the time taken between the parcel leaving the airplane and it striking the ground b)the horizontal distance travelled by the parcel in this time. c)the speed at which the parcel strikes the ground

Answer 1

a)The time taken between the parcel leaving the airplane, and it's striking the ground will be 6.38 second

c)The speed at which the parcel strikes the ground will be 62.64 m/sec.

What is speed?

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by U for the initial speed while V for the final speed. its SI unit is m/sec.

The given data in the problem is;

Steady speed (V)= 90 m/s

Altitude (h) = 200 m

Time taken (t)=?

horizontal distance travelled(d)=?

Striking speed(V)=?

The striking speed is;

$\rm V^2=U^2+2gh \\\\ V=\sqrt{2gh} \\\\V=\sqrt{2\times 9.81 \times 200} \\\\ V= \sqrt{3924} \\\\ V=62.64 \ m/sec$

The time taken is found as;

$\rm H = ut + \frac{1}{2} gt^2 \\\\ 200=0+ \frac{1}{2} \times 9.81 \times t^2 \\\\ t=6.38 \ sec$

Hence, the time taken between the parcel leaving the airplane, and it's striking the ground will be 6.38 second

To learn more about the sped, refer to the link;

https://brainly.com/question/7359669

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