A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m/s at an angle of 35.0 ∘ with the horizontal, as shown in the figure (Figure 1) . Determine the time taken by the projectile to hit point P at ground level.

9.96 s

Explanation:

Given in the y direction:

Δy = -115 m

v₀ = 65.0 m/s sin 35.0° = 37.28 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-115 = 37.28 t − 4.9 t²

4.9 t² − 37.28 t − 115 = 0

t = [ 37.28 ± √(37.28² − 4(4.9)(-115)) ] / 9.8

t = (37.28 ± 60.37) / 9.8

t = 9.96

report

explanation:

thermal energy is the answer