Someone please provide the correct answers along with the method od solution, th

Question

Someone please provide the correct answers along with the method od solution, that would help out alot. Thanks

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.40 m/s at an angle of 21.0° below the horizontal. It strikes the ground 6.00 s later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Explanation / Answer

PLEASE UPVOTE. ALSO, COMMENT BELOW FIRST IF YOU HAVE ANY DOUBT.

1.

Given that

Initial velocity = V0 = 8.40 m/sec at 21 deg below the horizontal.

V0x = V0*cos theta = 8.40*cos 21 deg = 7.84 m/sec

V0y = -V0*sin theta = -8.40*sin 21 deg = -3.01 m/sec

See that V0y will be negative as angle is below the horizontal.

Now time taken by ball to reach the ground = 6.00 sec

Since there is no acceleration in horizontal direction, ax = 0, So

Horizontal distance = V0x*t

Horizontal distance = 7.84*6 = 47.04 m

Part B.

Using 2nd kinematic equation

H = V0y*t + 0.5*ay*t^2

ay = vertical acceleration = -g = -9.81 m/sec^2

So,

H = -3.01*6.00 - 0.5*9.81*6.00^2

H = -194.64 m

Height of Building = 194.64 m

Part 3,

Again Using same equation

H = V0y*t + 0.5*ay*t^2

this time H = -10 m, and t = ?

-10 = -3.01*t - 0.5*9.81*t^2

4.905*t^2 + 3.01*t - 10 = 0

Solving above quadratic equation

t = [-3.01 +/- sqrt (3.01^2 + 4*4.905*10)]/(2*4.905)

by taking + ve sign

t = 1.15 sec

Please Upvote.

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