Answer:
104.3 cm or 179.7
Explanation:
First find time that it takes for the object to hit the ground
*
Then find xf of projectile 
not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7 or 142+37.7
hope that helps
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
A) the periodic time is given by the equation;
T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
= 6.28 × √0.0673
= 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
thus; frequency = 0.614 Hz
b) The vertical distance, the height is given by
h= 0.66 cos 12
h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
mgh = 1/2mv²
v² = 2gΔh = 2×9.81 × 0.01
= 0.1962
v = 0.443 m/s
c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
= 1/2 × 0.310 × 0.443²
= 0.0304 J