Distance the dog has to travel to catch a ball - projectile motion

[dani123]

Homework Statement

Jaime is outside playing with her dog, Skip. She throws a tennis ball at an angle of 60° with an initial speed of 10.2m/s and from a height of 196cm. If Skip was standing beside her when she threw the ball, how far would Skip have to run in order to catch the ball just the moment before it hits the ground?

Homework Equations

d_v=1/2*at²
d_h=V_h*Δt
Kinetic equation d=V_i*t+ 1/2*at²
d_h=-V²*sin2θ /g
sinθ=opp/hyp
cosθ=adj/hyp
Δt=-2Vsinθ / g

The Attempt at a Solution

d_v= sin(60)*10.2m/s= 8.83m

The actual vertical height for the dog is 1.96m+8.83m=10.8m

t=√[10.8m/(0.5)*(9.8)]=1.48s

Then d_v=1/2*at²= 25.9m

If someone could just verify my work and make sure my answers are correct that would be greatly appreciated. Also, if you could check my significant figures are being respected. Thanks so much in advance!

 

[hypnosam]

Why do you get that height for the height of the dog? -where does the 8.83m come from?

What I did is found the time using s=ut+1/2at^2, solved that, got two answers, one of them negative which isn't valid.
Using this equation 10.2*t*sin60 -4.9t^2 +1.96=0, and solving gives the time as 2.0025 seconds. Then multiply this by the speed of the ball, negating air resistance, to get the horizontal distance. This gives you 2.0025*10.2cos60 = 10.21275m.

 

[dani123]

Thank you so much for your time to look and address the issues i am having with this problem. I am just a little confused when it comes to the variables you used. What does the s represent? and same question goes for the u? If you clarify this with me that would be fantastic! thanks!

 

[hypnosam]

Oh, s is the distance, u is the initial speed, v is the final speed a as acceleration and t as time.

 

[dani123]

Thank you!

 

[dani123]

This is a really stupid question but what do you by you got two answers and one of them was negative?

 

[dani123]

I redid the question and got 10.21m as well! Thank you very much! :)