A tennis ball rolls off the edge of a tabletop of height 0.810 m above the floor and strikes the floor at a point a horizontal distance 1.70 m from the edge of the table. You can ignore air resistance.
Time of flight is: 0.407 s
Magnitude of initial velocity is: 4.18 m/s
Find the magnitude of the velocity of the ball just before it strikes the floor.
Take the free fall acceleration to be g = 9.80 m/s^2 .
Find the direction of the velocity of the ball just before it strikes the floor.
Express your answer as an angle measured from below the horizontal.
2007-02-06 21:50:23 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Physics
use v=u+gt......
2007-02-06 21:55:42 · answer #1 · answered by toby 2 · 0⤊ 0⤋
the velocity has 2 components
An horizontal component which does not change and is
VH =4.18 m/s (you can check this doing 4.18*0.407=1.7 distance made horizontally by the ball)
Averical component which is given by the formula V V = g*t here 0.404*9.8 = 4ms
So to find magnitude of velocity you have a rectangular triangle
with sides 4 and 4.18 and you search hypothenuse
V^2 = 4^2+4.18^2=33.47 and to find V = 33.47^0.5 = 5.79 m/s
To find the angle you have its tangent which is 4/4.18 =0.96
and yousearch with computer the angle 43.7°
2007-02-06 22:10:51 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
The vertical velocity when it hits the ground is
v = u + at.
Since u = 0 and a = 9.8 and t = 0.407 s
v = 3.99 m/s
Horizontal velocity is 4.18
Resultant velocity is √ ( 3.99 ^2 + 4.18^2)
= 5.78 m/s.
tan Ө = vertical speed / horizontal speed.
tan Ө = 3.99 / 4.18
Ө = 43.67 degree below thehorizontal.
2007-02-07 04:10:10 · answer #3 · answered by Pearlsawme 7 · 0⤊ 0⤋
a=(v-u)/t
9.8 x .407=v-4.18
V=8.1686 m/s
oviously the direction of velocity is downwards.there is no angle given that means the ball hits the floor and the gradient is 0.so the edge of the table and the point where the ball hits is on the same line.
2007-02-06 22:01:52 · answer #4 · answered by tonima 4 · 0⤊ 0⤋
Use v=u-gt for vertical direction
Take u=0 ,t=0.407 and get v(y) in vertical direction
Then use v=u for horizontal direction(because a=0)to get v(x)
and V= [v(x)^2 + v(y)^2]^1/2
If angle is 'A'(below horizontal)
then tan A=v(y) / v(x)
2007-02-06 22:43:04 · answer #5 · answered by neeraj_agarwal_1990 1 · 0⤊ 0⤋
Thrust eqn is T = mass flow rate of exhaust x exhaust speed = 1300 x 40000 = 5.2 x 10^7 N
2016-05-24 02:29:19 · answer #6 · answered by Anonymous · 0⤊ 0⤋