Answer:
Explanation:
The question relates to time of flight of a projectile .
Time of flight = 2 u sinθ / g
u is speed of projectile , θ is angle of projectile
= 2 x 48.5 sin42 / 9.8
= 6.6 seconds .
Maximum height attained
= u² sin²θ / g
= 48.5² sin²42 / 9.8
= 107.47 m .
Answer:
a
Explanation:
When people jump from greater height, energy gets absorbed into the muscles as potential. And when the person flexes her legs on landing energy is trans formed not only into elastic energy in the bones, but also into elastic energy in other tissues and into thermal energy.
air resistance is very small hence, negligible. So, the statement a is correct and rest are wrong.
Answer:
Using the range formula R = v^2 sin 2 theta / g
or v^2 = R * g / sin 86.4
v^2 = 3.14 m * 9.81 m/s2 / .998
v^2 = 30.9 m^2 / s^2
v = 5.56 m/s
This hasn't really proved the question - this would give
vy = 5.56 * sin 43.2 = 3.81 m/s
vx = 5.56 * cos 43.2. = 4.05 m/s
t = 1.57 / 4.05 = .387 sec to reach the waterfall
h = 3.81 * .387 - 4.9 (.387)^2 = .74 m well above the height of the falls
There seems another way to do this
vy / vx = tan 43.2 vy = .939 vx
h = vy t - 1/2 g t^2 and t = 1.57 / vx
h = 1.57 tan 43.2 - 4.9 (1.57 / vx)^2
Solving for vx I get vx = 3.26 m/s vy = 3.06 m/s v = 4.47 m/s
Recall the definitions of
• average velocity:
v[ave] = ∆x/∆t = (x[final] - x[initial])/t
Take the initial position to be the origin, so x[initial] = 0, and we simply write x[final] = s. So
v[ave] = s/t
• average acceleration:
a[ave] = ∆v/∆t = (v[final] - v[initial])/t
Assume acceleration is constant (a[ave] = a). Let v[initial] = u and v[final] = v, so that
a = (v - u)/t
Under constant acceleration, the average velocity is also given by
v[ave] = (v[final] + v[initial])/2 = (v + u)/2
Then
v[ave] = s/t = (v + u)/2 ⇒ s = (v + u) t/2
and
a = (v - u)/t ⇒ v = u + at
so that
s = ((u + at) + u) t/2
s = (2u + at) t/2
s = ut + 1/2 at²
