Solving A problem using kinematics and Energy

[lion_]

Homework Statement

A $1.50$ kg snowball is fired from a cliff $12.5$m high with an initial velocity of $14.0$ m/s, directed $41^{\circ}$ above the horizontal. How fast does the ball travel when it hits the ground.

Homework Equations

$E=K+U$
$V^2_f=v^2_i+2a(y_f-y_i)$
$y_f=v_i \sin \theta-gt^2$

The Attempt at a Solution

First using conservation of mechanical energy,
$E=(1/2mv_f^2-1/2mv^2_i)+(0-mgh)$
$v=\sqrt{v^2_i+2mgh} \implies v=\sqrt{14^2+2((9.8)(12.5)}=21m/s$

Using kinematics to verify,
$0=v_i\sin \theta -gt$
$t=\frac{v_i\sin \theta}{g}=\frac{14\sin 41}{9.8}=0.937 s$
$y_f=14\sin 41 (.937)-.5(9.8)(.937)^2=4.3m$ which gives the height traveled above the horizontal
Calculate speed $v=\sqrt{2(9.8)(12.5+4.3)}=18.14m/s$

Which is wrong they both look correct.

 

[SammyS]

Homework Statement

A $1.50$ kg snowball is fired from a cliff $12.5$m high with an initial velocity of $14.0$ m/s, directed $41^{\circ}$ above the horizontal. How fast does the ball travel when it hits the ground.

Homework Equations

$E=K+U$
$V^2_f=v^2_i+2a(y_f-y_i)$
$y_f=v_i \sin \theta-gt^2$

The Attempt at a Solution

First using conservation of mechanical energy,
$E=(1/2mv_f^2-1/2mv^2_i)+(0-mgh)$
$v=\sqrt{v^2_i+2mgh} \implies v=\sqrt{14^2+2((9.8)(12.5)}=21m/s$

Using kinematics to verify,
$0=v_i\sin \theta -gt$
$t=\frac{v_i\sin \theta}{g}=\frac{14\sin 41}{9.8}=0.937 s$
$y_f=14\sin 41 (.937)-.5(9.8)(.937)^2=4.3m$ which gives the height traveled above the horizontal
Calculate speed $v=\sqrt{2(9.8)(12.5+4.3)}=18.14m/s$

Which is wrong they both look correct.

The first method looks correct.

In the second method, you completely ignored the horizontal component of the velocity.

 

[lion_]

How is the horizontal component needed? The only thing we are concerned with is the height, thus once the snowball reaches maximum height, it becomes a one dimensional problem regardless of how fast the horizontal component is traveling. It is independent of gravity.

 

[SammyS]

How is the horizontal component needed? The only thing we are concerned with is the height, thus once the snowball reaches maximum height, it becomes a one dimensional problem regardless of how fast the horizontal component is traveling. It is independent of gravity.

How is it not needed?

Aren't you asked:

How fast does the ball travel when it hits the ground.

How is speed related to velocity?

 

[lion_]

So you need to calculate $v_f=\sqrt{v_x^2+v_y^2}=\sqrt{14^2\cos^2 41^{\circ}+(18.14)^2}=20.99$ I guess that what was missing thanks.