Physics...help!? About throwing a ball at an angle?

Question

Can you explain how to do this?

1. A baseball was thrown from 1 m hill. Assume that the ball had an initial velocity of 70 km/h (19.44 m/s) at 40 degrees.

c. How long will it take for the ball to land on the ground?

d. How far will the ball travel?

Thanks

Answer 1

horizontal velocity: Vh = V*cos(40)

=> Vh = 14.89m/s

vertical velocity: Vv = V*sin(40) - g*t

=> Vv =12.50 -9.8*t

horizontal position: x = x0 + vx0t  

=> x =14.89*t

vertical position: y = y0 + vy0t - 1/2gt²

=> y= 1 +12.50*t - 0.5gt²

when the ball hit the ground y = 0

=> 1 +12.5*t- 4.9*t² = 0

solve for t and take the positive result since time can't be negative.

t=2.63 seconds.

c. It'll take 2.63 seconds

d. x = 14.89*2.63 = 39.1m

the ball will travel 39.1m

Answer 2

Only 1 m hill?

The ball will reach it max height h at

t1= Vv/g

where Vv is the vertical component  of V

Vv= V sin(40)= 19.44  sin(40)=12.5 m/s

then  t1= 1.27s

The max height above the hill is h

h= 0.5 g t^2

h= 0.5 g ( Vv /g)^2= 0.5 ( Vv )^2 /g

h= 0.5 (12.5)^2 / 9.81= 8.0 m

then it will fall the from the height  H

H= h + 1

that would take

t2= sqrt( 2H/g)=sqrt( 2 x 9 /9.81)=1.35 s

Finally total time is

t= t1+ t2

t= 1.27 + 1.35= 2.62 s

Now since you know how long it was traveling

S=Vh t

where Vh = V cos(40)= 19.4 cos(40)=14.9 m/s

S= 14.9 x 2.63 = 39 m