A snowball is thrown with an initial x velocity of 7.5 m/s and an initial y velocity of 8.4 m/s . Part A How much time is requir
1 answer:
In order to calculate the time taken by the snowball to reach the highest point in its journey, we need to consider the variables along the y-direction.
Let us list out what we know from the question so that we can decide on the equation to be used.
We know that Initial Y Velocity = 8.4 m/s
Acceleration in the Y direction = -9.8 m/
, since the acceleration due to gravity points in the downward direction.
Final Y Velocity = 0 because at the highest point in its path, an object comes to rest momentarily before falling down.
Time taken t = ?
From the list above, it is easy to see that the equation that best suits our purpose here is
Plugging in the numbers, we get 0 = 8.4 - (9.8)t
Solving for t, we get t = 0.857 s
Therefore, the snowball takes 0.86 seconds to reach its highest point.
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Answer:
987 joules, 3.01s
Explanation:
(A)
from the attached diagram
net force, Fnet, pulling the crate up the ramp is given by
Fnet = FcosФ - WsinФ - Fr
where FcosФ is the component of horizontal force 290N resolved parallel to the plane
WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane
Fr = constant opposing frictional force
Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65
Fnet = 240.421 - 109.602 - 65
Fnet = 65.82N
Work done on the crate up the ramp, W, is given by
W = Fnet × d (distance up the plane)
W = 65.819 × 15
W = 987.285 joules
W = 987 joules (to 3 significant Figures)
(B)
to calculate the time of travel up the ramp
we use the equation of motion
where s = distance up the plane, 15m
u = Initial velocity of the crate, which is 0 for a body that is initially at rest
a = acceleration up the plane, given by
where m = mass of the crate, 20 kg
now,
from,
15 = 0 + 1.645
15 = 1.645
t = 3.01s (to 3 sig fig)
