Solving Questions on Throwing a Baseball to Hit a Target

[Sassenav22]

1. Homework Statement

You throw a baseball to hit a target on the ground 10m from the base of a building 25m in height. (a) With what velocity must you throw the baseball if it is to leave the hand horizontally? (b) with what velocity mut you throw the baseball if it is to leave the hand at a angle of 45 up from the horizontal? (c0 What is the horizontal component of the initial value o the velocity in case (b)?

Homework Equations

(a) s=ut +1/2at^2
v=d/t

(b) i have no clue

(c) Vx(cos theta)

The Attempt at a Solution

(a)s=ut+1/2at^2
25=0+ 1/2(9.8)t^2
25= 4.9t^2
25/4.9=t^2
t=√5.10
t=2.26s

v=d/t
= 10/2.26s
=4.42m/s

(b) i don't know can someone point me in the right direction

 

[LowlyPion]

If it is thrown at 45° then Vx = Vy.

That means you can take your x and y equations and solve more easiily.

25 = Vy*t - ½gt²

10 = Vx*t

With Vx = Vy and t = 10/V ...

 

[Sassenav22]

so its:
25=4.42(cos45)(t)+0.5(9.8)t^2

and solve for t

then use v=d/t

and find the velocity?

 

[LowlyPion]

I'd just stick 10/V into the y equation for t and solve for Vx,Vy directly.

Then Vo = V/sin45 = Vx(√ 2)

 

[Sassenav22]

i am confused

 

[LowlyPion]

Since -25 = Vy*t - ½gt²

And |Vx| = |Vy| and

Vx = 10/t

Then Vy = 10/t or t = 10/Vy

Hence -25 = Vy*10/Vy - ½*g*(10/Vy)²

Simplifying 70V² = 100*g = 980

Vy = Vx = 3.74 m/s

Vo = 5.29 m/s