Question

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0 degrees below the horizontal. It strikes the ground 3.00s 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?

Answer 1

can anyone try to solve this. please

Answer 2

a) the horizontal speed of the ball is constant since there are no horizontal forces acting; the horizontal speed of the ball is 8.4cos25=
7.6 m/s

once we find the time the ball is in flight, we know the horizontal distance is 7.6m/s x time of flight = 7.6m/s x 3x = 22.8 m

b) we can use the equation of motion:

y(t)=y0+v0yt+1/2at^2 where y is the height at any time, t, y0 is initial height (what we are solving for), v0y is the initial vertical speed and a is accel

the initial vertical speed = -8.4sin25 = -3.6m/s; a = g = -9.8 m/s/s

the negative signs are used since these vectors point down

we know that y is zero when t=3, giving us:

0=y0-3.55m/s x 3x -4.9m/s/s x 9 s^2

y0=10.65+44.1=54.75 m

c) the object will travel a distance of 10 m down in a time given by:

dist = v0yt+1/2gt^2

or 10=3.55t+4.9t^2

this is a quadratic equation:

4.9t^2+3.55t-10=0
whose positive solution is t=1.1s