At a particular instant, a stationary observer on the ground sees a package fall

Question

At a particular instant, a stationary observer on the ground sees a package falling with speed v_1 at an angle to the vertical. To a pilot flying horizontally at constant speed relative to the ground, the package appears to be falling vertically with a speed v_2 at that instant. What is the speed of the pilot relative to the ground? v_1 + v_2 v_1 - v_2 v_2 - v_1 Squareroot v_1^2 - v_2^2 Squareroot v_1^2 + v_2^2 An object released from rest at time t = 0 slides down a frictionless incline a distance of 1 meter during the first second. The distance traveled by the object during the time interval from t = 1 second to t = 2 seconds is 1m 2m 3m 4m 5 m Two people are in a boat that is capable of a maximum speed of 5 kilometers per hour in still water, and wish to cross a river 1 kilometer wide to a point directly across from their starting point. If the speed of the water in the river is 5 kilometers per hour, how much time is required for the crossing? 0.05 hr 0.1 hr 1 hr 10 hr The point directly across from the starting point cannot be reached under these conditions. A projectile is fired from the surface of the Earth with a speed of 200 meters per second at an angle of 30 degree above the horizontal. If the ground is level, what is the maximum height reached by the projectile? 5 m 10 m 500m 1.000 m 2.000 m

Explanation / Answer

12. suppose angle is theta then

Vpackage = v1 sin(theta)i - v1 cos(theta)j

Suppose pilot is flying with velocity, Vpilot = Vpi

Vpackage wrt Vpilot = Vpackage - Vpilot

-v2j = (v1 sin(theta) - Vp) i - v1 cos(theta)j

henece v1 cos(theta) = v2

and v1 sin(theta) = vp

square and add

v1^2 (sin^2 + cos^2) = v2^2 + vp^2

vp = sqrt(v1^2 - v2^2)

13. d = ut + at^2 / 2

and u = 0 (initial velocity)

d = a t^2 /2

for t = 0 to 1 sec/

d1 = 0.5a

from 0 to 2 sec

d2 = a(2^2) / 2 = 2a

during 1 to 2 sec, d = d2 - d1 = 1.5 a

and d1 = 0.5a = 1

a = 2

so d = 1.5 x 2 = 3 m

14. if you want to reach at directly across point.

then you have to have a component of velocity of boat in opposite direction of river flow to cancel its effect.

but Vriver = 5 km/h

and all speed you have is 5 km/h.

if you will sail in opposite diretion then you will at that point all the time

(you are not moving)

and if you sail in any other direction , then you will flow with river hence you can never reach the directly across point.

15. velocity in vertical = 200 sin30 = 100 m/s

vfy^2 - viy^2 = 2 a d

0^2 - (100^2) = 2(-9.8)(H)

H = 510 m

Ans(C)

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