How Far Does the Ball Travel Horizontally Before Bouncing?

[kman2027]

A ball is thrown with an initial speed of 4.6m/s at an angle of 15(degrees) below the horizontal. It is released 0.80m above the floor.

What horizontal distance does the ball cover before bouncing?



I keep getting 2.4m as my answer, but it's wrong!
Here's what I did (Please correct any mistakes):


4.6sin(15) = initial vertical velocity = 1.19m/s

1.19^2 = 2gh

1.42 = 2gh

1.42/(2*9.81)=h= .072m

upward time = 1.19 m/s - gt = 0
1.19/9.81 = t = .121sec

total height = .072 + 0.8 = .872m

0.872 = 1/2gt^2
0.872/4.9 = t^2 = 0.178, sqrt = t = 0.422sec

total time = 0.121 + 0.422= 0.543sec

horizontal velocity = 4.6cos(15) = 4.44m/s

4.44x 0.543 = 2.41m (answer)

 

[tiny-tim]

welcome to pf!

hi kman2027! welcome to pf!

1.19^2 = 2gh

sorry, but this is weird …

energy is a scalar, you can't split it into components in different directions!​

use one of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations, with v_i = 1.19

 

[NewtonianAlch]

Remember that gravity acts on the object as soon as it is released, so the time it is in the air is dependent on that, regardless of how fast you're throwing it in the horizontal direction.

 

[kman2027]

A ball is thrown with an initial speed of 4.6m/s at an angle of 15(degrees) below the horizontal. It is released 0.80m above the floor.

What horizontal distance does the ball cover before bouncing?

Is the answer 2.98?

Here's what I did:
4.6sin(15) = initial vertical velocity = 1.19m/s
horizontal velocity = 4.6cos(15) = 4.44m/s


.8/(1.19m/s)=.672s
0+(4.44m/s)(.672s)=2.98

 

[tiny-tim]

hi kman2027!

.8/(1.19m/s)=.672s
0+(4.44m/s)(.672s)=2.98

the first equation assumes that the vertical speed is constant (s = ut), it isn't

you need to use one of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations to find t

(and then your method in the second equation is correct, since the horizontal speed is constant )

 

[kman2027]

I got 1.7m this time.

I picked the constant acceleration equation:

y=yo+voyt-1/2gt^2
plugged in 0=.8+1.19t-1/2(-9.8)t^2
t=.385

(.385)(4.44)=1.7m

Is 1.7m right?

 

[tiny-tim]

hi kman2027!

(just got up :zzz: …)

plugged in 0=.8+1.19t-1/2(-9.8)t^2

not quite …

the initial vertical speed is negative …

(and you shouldn't have both those minuses in the t² term)

A ball is thrown with an initial speed of 4.6m/s at an angle of 15(degrees) below the horizontal.

 

[kman2027]

My bad, I had a typo in the equation.

.8-1.19t+(1/2)(-9.8)t^2=0
t= .3s

Then:
(4.44)(.3)= 1.3m

Is 1.3m it?

Thanks for your help by the way!

 

[NewtonianAlch]

My bad, I had a typo in the equation.

.8-1.19t+(1/2)(-9.8)t^2=0
t= .3s

Then:
(4.44)(.3)= 1.3m

Is 1.3m it?

Thanks for your help by the way!

What's the equation for calculating distance based on velocity and time?

 

[tiny-tim]

yes 1.3 looks fine