Answer:
a) v₀x = 9.05 m/s (→)
b) v₀y = 33.63 m/s (↑)
Explanation:
Given
t = 2.1 s
x = 19 m
y = 49 m
a) v₀x = ?
We can apply
x = vx*t = v₀x*t ⇒ v₀x = x/t = (19 m) / (2.1 s) = 9.05 m/s
⇒ v₀x = 9.05 m/s (→)
b) v₀y = ?
Let's use this equation
y = y₀ + v₀y*t - 0.5*g*t²
⇒ 49 m = 0 m + v₀y*(2.1 s) - 0.5*(9.81 m/s²)*(2.1 s)²
⇒ v₀y = 33.63 m/s (↑)
If we get ymax as follows
ymax = v₀y² / (2*g) = (33.63 m/s)² / (2*9.81 m/s²) = 57.66 m
we can see that 57.66 m > 49 m (ymax > y)
