iPad 11:51 PM 1%* Homework: Vectors And 2.D Kinematics Catch Mie throws ball to

Question

iPad 11:51 PM 1%* Homework: Vectors And 2.D Kinematics Catch Mie throws ball to her friend Sarah Thhe ball leaves Jae's hand distance,5 meters above the ground with an nitial speed of 12 m/s at an angle 4 degreet with respect to the horizontal, Sarah catches the ball 1.5 meters above the ground. What is the horinontal component of the ball's velocity when it leaves Jubie's hand 8.023 Tipler6 1.P.054. 2, what isth, vertical component of the bar velocity when it leaves Am, hand, 8.9177 Tiplers 3.P.007 Tipler6 3 P.010. 3) What is the maxinum height the ball goes above the ground Tiplers 3.P.023 Tiplers 3.P.047 What is the distarce between the two girls Tiplers 3.P.072 Tiplers 3.P.109 After canching the ball, Sarah throws it back to Jutie. The bal leaves Sarah's hand a distance 1.5 meters above the ground.and is mong with speed of10m/s when t reaches a manmum height of 7 m above te gound. What is the speed of the ball when it leaves Sarah's hand 6) How high above the ground will the ball be when it gets to Jie inote, the ball may go ever Jdie's head 7) Below is some space to write notes on this problem

Explanation / Answer

We have given following details

Initina velocity say u=12m/s

Angle of projection say x=45°

Highet of man above ground=1.5m

So we have

1):the velocity of ball can be divided into 2 components.one along horizontal and one along vertical

Along horizontal we have, velocity=ucosx

I.e velocity x cosine of angle

So along x axis we have, ucosx=12*cos(45°)

=12*1/2=12/2 m/s. =

2); along vertical direction we have vertical component = velocity * sine of angle

So vertical component= usinx

12* sin(45°)

12*1/2

12/2 m/s.

3): max highet of projection will be at the point where the vertical component of velocity become zero.

Using equation of motion

V² = U + 2as where v is final velocity

U is initial velocity in vertical i.e usinx =12/2

V is final velocity=0 as ball stops

a is acceleration due to gravity,i.e 9.8 m/s² or approximately 10m/s² which is opposite to motion of ball.so it will be negative.

And s is highet.

So solving above equation we get

0² = (12/2)² + 2(-10)(s)

144/2= 20 x s

S= 72/20

S= 3.6 m

So total highet will be equal to highet,s + initial highet i.e 1.5m

So H= total highet=3.6m + 1.5m = 4.1m above the ground.

4): distance between two girls= range of projectile

Horizontal velocity x time of flight

In simplifying the two formulas, we get range= V²sin2x/g

12² * sin(2*45°)/10

144* 1/10 as sin90= 1 and g= 10m/s²

144/10 = 14.4 m

5):At reaching the highest point there is no effect on horizontal component of velocity,as it remains constant throughout since the acceleration is acting only in vertical direction.so horizontal velocity at highest point is equal to initial horizontal velocity

=>Ucosx=7

10* cosx=7

Cosx=7/10

X,i.e angle of projection=cos-¹(7/10) degrees.which is approximately 45.573°.

6):here range of ball or the horizontal distance which ball covers=v²sin2x/g

Range=100*sin(2*45)/10

Range=100*1/10 as sin 90=1

Range=100/10

Range=10m.

As Julie is at the distance of 14.4 m,so ball will not reach to Julie.it wall fall down to ground before reaching Julie.

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