Motion.

A parachutist after bailing out falls 50m without friction. When the parachute opens, it decelerates at 2ms−2. He reaches the ground with speed 3ms−2. At what height did he bail out

Expert Solution

The fundamental concepts of motion of any object are described by the basic equations of motion. These equations control an object’s motion in one, two, and three dimensions. These equations are used to calculate the components of any object, such as velocities, displacement, and acceleration. As a result, they can only be employed when motion is in a straight line with constant acceleration.

Initial velocity,u = 2gh

or,u = √2 × 9.8 × 50= 145

The velocity at ground,v = 3m/s

v2 − u2 = 2as

or,32 − 980 = 2×(−2)×s

Thus, s = 971/4,which is nearly 243 m.

Initially, he has fallen 50 m

Thus, total height at which he bailed out = 243 + 50 = 293m

Total height at which he bailed out = 293m

Nikhil Thakur answered 2 years ago

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