Question

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00m/s at an angle of 20.0 below the horizontal. It strikes the ground 3.00s later. How long does it take the ball to reach a point 10.0m below the level of launching?

Answer 1

=1.18s

Step-by-step explanation:

(a) x

f

​

=v

xi

​

t=(8.00m/s)cos20.0

0

(3.00s)=22.6m

(b) Taking y positive downwards,

y

f

​

=v

yi

​

t+

2

1

​

gt

2

y

f

​

=(8.00m/s)sin20.0

0

(3.00s)+

2

1

​

(9.80m/s

2

)(3.00s)

2

=52.3m

(c) 10.0m=(8.00m/s)(sin20.0

0

)t+

2

1

​

(9.80m/s

2

)t

2

suppressing units

4.90t

2

+2.74t−10.0=0

t=

9.80

−2.74±

(2.74)

2

+196

​

​

=1.18s