The acceleration due to earth's gravity is -9.8 m/s [dn] I thought... I'm assuming this is a projectile motion question asking for the range.
Break each kinematic quantity into their x and y components.
x y
v₁ = 3 m/s 0
v₂ = 3 m/s ?
Δd = ? -1.5
Δt = ? ?
a = 0 -10 m/s²
So the variable we are trying to find is Δdx (x component of displacement). We need to use a kinematic equation to do so. However, we obviously don't have enough given to find Δdx. This means we need to find something first, something we can use. How about Δt? Δt can be applied to both the x and y components. We have enough information in the y component list to find Δt. We can use this formula and solve for Δt.
Δdy = v₁y ( Δt ) + 1/2 ( ay ) ( Δt )²
Δdy = 1/2 ( ay ) ( Δt )² <- the first term cancels out since v₁y = 0.
2Δdy = ay ( Δt )²
2Δdy / ay = ( Δt )²
√ 2Δdy / ay = Δt
√ 2(-1.5 m/s) / -9.8 m/s² = Δt
√ -3.0 m/s / -9.8 m/s² = Δt
√ 0.3<u>0</u>6122449 s² = Δt
0.5<u>5</u>32833352 s = Δt
Now, we can use this newly found quantity to solve for Δdx using the x component values using the appropriate kinematic equation.
Δdx = ( v₁x + v₂x / 2) ( Δt )
Δdx = ( ( 3.0 m/s + 3.0 m/s ) / 2 ) ( 0.5<u>5</u>32833352 s )
Δdx = ( 6.0 m/s / 2 ) ( 0.5<u>5</u>32833352 s )
Δdx = ( 3.0 m / s )( 0.5<u>5</u>32833352 s )
Δdx = 1.<u>6</u>59850006 m
Therefore, the girl should place her cage 1.7 m away from the platform to catch the lizard.
This solution assumes that the acceleration due to gravity is -10 m/s² [dn] and not -9.8 m/s² [dn]. If you need -9.8 m/s² [dn], then just substitute it into my solution. This was a pain to type lol
