What have I done wrong? (projectile motion)

[Mayzu]

Homework Statement

A cricketer fields the ball in the outfield, some 80 meters from the wicket keeper. The fielder needs to return the ball to the wicket keeper as quickly as possible. He can throw the ball with a speed of 140 km/hr. The cricket ball has a mass of 0.0168kg
(e) The fielder decides to throw the ball straight back to the wicket keeper, 80 m away. At what angle to the horizontal must he project the ball so that the ball reaches the wicket keeper without hitting the ground?

This question is worth 8 marks, but I cannot figure out how to do it!

Homework Equations

Vf=Vi+a(delta)t

The Attempt at a Solution

http://imgur.com/OJEDalz

^here's my working out. I just can't get past the last step, and my graphics calculator can't either, making me think I'm doing it completely wrong.

 

[berkeman]

Homework Statement

A cricketer fields the ball in the outfield, some 80 meters from the wicket keeper. The fielder needs to return the ball to the wicket keeper as quickly as possible. He can throw the ball with a speed of 140 km/hr. The cricket ball has a mass of 0.0168kg
(e) The fielder decides to throw the ball straight back to the wicket keeper, 80 m away. At what angle to the horizontal must he project the ball so that the ball reaches the wicket keeper without hitting the ground?

This question is worth 8 marks, but I cannot figure out how to do it!

Homework Equations

Vf=Vi+a(delta)t

The Attempt at a Solution

http://imgur.com/OJEDalz

^here's my working out. I just can't get past the last step, and my graphics calculator can't either, making me think I'm doing it completely wrong.

Where does the number 38.9 come from in your work?

I think the right approach is to key in on Vy=0 at 40m, and use the time from that equation to help the rest of your calculations...

 

[Mayzu]

Where does the number 38.9 come from in your work?

I think the right approach is to key in on Vy=0 at 40m, and use the time from that equation to help the rest of your calculations...

38.9 is the initial velocity of the ball. That's what I've done, but I'm stuck...

 

[cnh1995]

Homework Statement

A cricketer fields the ball in the outfield, some 80 meters from the wicket keeper. The fielder needs to return the ball to the wicket keeper as quickly as possible. He can throw the ball with a speed of 140 km/hr. The cricket ball has a mass of 0.0168kg
(e) The fielder decides to throw the ball straight back to the wicket keeper, 80 m away. At what angle to the horizontal must he project the ball so that the ball reaches the wicket keeper without hitting the ground?

This question is worth 8 marks, but I cannot figure out how to do it!

Homework Equations

Vf=Vi+a(delta)t

The Attempt at a Solution

http://imgur.com/OJEDalz

^here's my working out. I just can't get past the last step, and my graphics calculator can't either, making me think I'm doing it completely wrong.

Are both of them (the players) assumed to be just two points? Shouldn't their heights be taken into account? In that case, you could use 38.9 as initial velocity (resultant) and put that in the range formula to get sin2θ. That value of θ would be the critical one, means just enough to make the ball reach the ground.

 

[cnh1995]

Homework Statement

A cricketer fields the ball in the outfield, some 80 meters from the wicket keeper. The fielder needs to return the ball to the wicket keeper as quickly as possible. He can throw the ball with a speed of 140 km/hr. The cricket ball has a mass of 0.0168kg
(e) The fielder decides to throw the ball straight back to the wicket keeper, 80 m away. At what angle to the horizontal must he project the ball so that the ball reaches the wicket keeper without hitting the ground?

This question is worth 8 marks, but I cannot figure out how to do it!

Homework Equations

Vf=Vi+a(delta)t

The Attempt at a Solution

http://imgur.com/OJEDalz

^here's my working out. I just can't get past the last step, and my graphics calculator can't either, making me think I'm doing it completely wrong.

If your calculations are right so far, you're just a step away from the answer. sin2θ is 2sinθcosθ..