- #1

- 322

- 45

- Homework Statement
- Determine the minimum initial velocity v0 at which the ball must be kicked in order for it to just cross over the 3-m high fence .

Part B:

Determine the corresponding angle ##\theta _0## at which the ball must be kicked.

- Relevant Equations
- kinematic eqs for constant acceleration

Hello there, I don't understand what I'm doing wrong I don't get the correct answer, but have done the same analysis 3x already and still get the same...

Some input would be appreciated thanks in advance.

Note: y-axis is upwards and x-axis is to the right.

3 unkowns i.e. 3eqs.

##x = x_0 + v_{0,x}t##

##6 = v_0cos\theta * t (1)##

##y = y_0 + v_{0,y}t + \frac 12 a_ct^2##

## 3 = v_0sin\theta * t - \frac12 gt^2 (2)##

##v_y = v_{0,y} - gt##

##0 = v_0cos\theta - gt (3)##

##(3) in (2)## --> ##3 = gt^2 - \frac 12 gt^2##

##\frac 12 gt^2 = 3## --> ##t = \sqrt{\frac 6g} = 0.7821s##

##\frac {(3)}{(1)} --> \frac {v_0sin\theta}{v_0cos\theta * t} = \frac {gt}{6}##

##tan\theta = \frac {gt^2}{6}##

##\theta = atan\frac {gt^2}{6}##

##\theta = 45## degrees -->

using these result in ##(1)##:

##v_0 = \frac {6}{cos\theta * t} = 10.85m/s##

Thanks in advance.

Some input would be appreciated thanks in advance.

Note: y-axis is upwards and x-axis is to the right.

3 unkowns i.e. 3eqs.

##x = x_0 + v_{0,x}t##

##6 = v_0cos\theta * t (1)##

##y = y_0 + v_{0,y}t + \frac 12 a_ct^2##

## 3 = v_0sin\theta * t - \frac12 gt^2 (2)##

##v_y = v_{0,y} - gt##

##0 = v_0cos\theta - gt (3)##

##(3) in (2)## --> ##3 = gt^2 - \frac 12 gt^2##

##\frac 12 gt^2 = 3## --> ##t = \sqrt{\frac 6g} = 0.7821s##

##\frac {(3)}{(1)} --> \frac {v_0sin\theta}{v_0cos\theta * t} = \frac {gt}{6}##

##tan\theta = \frac {gt^2}{6}##

##\theta = atan\frac {gt^2}{6}##

##\theta = 45## degrees -->

**WRONG**using these result in ##(1)##:

##v_0 = \frac {6}{cos\theta * t} = 10.85m/s##

**--> also WRONG because of the angle of course.**Thanks in advance.