Aprojectile is fired at an upward angle of 29.7â° from the top of a 108-m-high cliff with a speed of 130-m/s. what will be its speed (in m/s) when it strikes the ground below?

79.2 m/s

Explanation:

θ = angle at which projectile is launched = 29.7 deg

a = initial speed of launch = 130 m/s

Consider the motion along the vertical direction

v₀ = initial velocity along the vertical direction = a Sinθ = 130 Sin29.7 = 64.4 m/s

y = vertical displacement = - 108 m

a = acceleration = - 9.8 m/s²

v = final speed as it strikes the ground

Using the kinematics equation

v² = v₀² + 2 a y

v² = 64.4² + 2 (-9.8) (-108)

v = 79.2 m/s

real, smaller and upside-down.   so the answer should be b.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000