d. 49.0 m/s
<em><u>this</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>your</u></em><em><u> </u></em><em><u>answer</u></em><em><u>. </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>OK</u></em><em><u>. </u></em><em><u>.</u></em><em><u>.</u></em>
<em><u>please</u></em><em><u> </u></em><em><u>mark</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>brainliest</u></em><em><u>. </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
<em><u>follow</u></em><em><u> </u></em><em><u>me</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>my</u></em><em><u> </u></em><em><u>friend</u></em><em><u>. </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
The equation to be used here is the trajectory of a projectile as written below:
y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity
Since the angle is below horizontal, let's use the minus equation. Substituting the values:
- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m
However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.
Answer:
Sherman is right about the green used to camoflauge the dragonflies.
He is wrong about them being the **same** dragonflies.
Answer:
Bend to extend. Bending with a hips back movement, back straight, feet flat and forward. ...
Squat. This is a hips down motion. ...
Lunge. This is a long, linear stride, lowering your back knee to just above the ground, with a completely upright torso. ...
Rotate. ...
Push. ...
Pull. ...
Gait.
Answer:
2.5E24/4.168351075775951e
=4.168351075775951e
