need help in 2 physics questions asap?

Question

Question #1:
A clam dropped by a sea gull takes 3.0 seconds to hit the ground. What is the sea gull's approximate height above the ground at the time the clam was dropped?


Question #2:
An object is allowed to fall freely near the surface of a planet. The object falls 54 meters in the first 3.0 seconds after it is released. What is the acceleration due to gravity on that planet?

Answer 1

1) y = y_0 + v_0t + (1/2)a t^2
y = 0, v_0 = 0, a = -9.81 m/s^2, t = 3, we're solving for y_0
Rearrange the equation:

y_0 = (1/2)(9.81)(3^2) = 44.145 meters.

2) Using the same equation as the last one.
y = 0, y_0 = 54 meters, t = 3, v_0 = 0, we're solving for a
0 = 54 + (1/2)a(3^2)
Rearrange.
-54(2)/9 = a = -12 m/s^2

Answer 2

Apply s = ut + 1/2 * a * t^2 for both questions. s = height released, u=0 is the initial vel, a is the accel due to gravity, and t is the time of flight. For question 1, a=g is given, use t to solve s. For question 2, a is unknown but s and t are given.

Answer 3

Q#1:
formula: Newtons Law

Distance = (1/2)gt^2 , known t=3.0 secs.

where s = distance
g = gravitational acceleration = 9.81 m/s^2

t = time = seconds

Solve for Distance: s = (1/2)(9.81 m/s^2)(3.0 secs)^2

s = 44.145 meters

Answer 4

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Answer 5

2*height = a*t^2
Just plug in the data. works for both problems.