A ball is projected horizontally from the edge of a table that is 1.03 m high, a

Question

A ball is projected horizontally from the edge of a table that is 1.03 m high, and it strikes the floor at a point 1.34 m from the base of the table. What is the initial speed of the ball in m/s) D: 6.141 E: 8.91| F: 1.29×101| G: 1.87x101| A: 2.02|| B: 2.92|| C: 4.241 Submit Answer Tries 0/3 H: 2.72x101 How high is the ball above the floor when its velocity vector makes a 49.3° angle with the horizontal? (in m) A: 1.06x10-11 B: 1.41×10-11 Submit Answer Tries 0/3 C: 1 .88×10-1 1 D: 2.50×10-11 E: 3.32x10 E: 3.32x10 F: 4.42x101 G: 5.88x10-1 OH: 7.82x10-1

Explanation / Answer

(a)initial velocity along vertical =Voy=0

yo=1.03m

y=0

y=yo+Vot + 1/2 at2

0=1.03+0(t)+1/2(-9.8)(t)2

t=0.46 s

now distance along horizontal=X=1.34m

x= Vo (t)

1.34=Vo(0.46)

Vo=2.92 m/s

option (b) is correct.

(b) when theta=49.3

tan49.3= -Vy/Vx

Vy=2.92tan49.3

Vy= 3.4m/s

now Vy2=V2oy+2gh

(3.4)2=0+2 (9.8)h

h= 0.5898m

h is the height coverred from the top of the table

so the height above the floor=1.03-0.5898

H=0.442m= 4.42x10-1m

So option (f) is correct.

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