A squirrel jumps into the air with a velocity of 7 m/s at an angle of 20°. What is the maximum height reached by the squirrel?

I have my answer as 29.2 Am I right?

Anyone wanna check my answer?

no your answer is incorrect

To find the maximum height reached by the squirrel, we can use the kinematic equation for vertical motion:

Δy = v₀²sin²θ / (2g)

Where:
Δy = the change in height (maximum height reached)
v₀ = initial velocity (7 m/s)
θ = launch angle (20°)
g = acceleration due to gravity (9.8 m/s²)

Plugging in the values, we have:

Δy = (7 m/s)² * sin²(20°) / (2 * 9.8 m/s²)

Now, let's calculate this:

Δy = 49 m²/s² * sin²(20°) / 19.6 m/s²
Δy = 49 m²/s² * 0.139 / 19.6 m/s²
Δy ≈ 0.342 m

Therefore, the maximum height reached by the squirrel is approximately 0.342 meters.

Your answer of 29.2 Am is not correct. The units for height should be in meters (m), not Am (ampere-meter) which is a unit for electrical current.