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arsen [322]
3 years ago
15

A marshmallow is dropped from a 5.71 meter high pedestrian bridge, and 0.921 seconds later, it lands right on the head of an uns

uspecting person walking underneath. How tall is the person with the marshmallow on her head?
a: 1.75 m
b: 1.70 m
c: 1.66 m
d: 1.51 m
e: 1.55 m
f: 1.89 m
g: 1.49 m
h: 1.62 m
Physics
1 answer:
Natalka [10]3 years ago
5 0

let the height of the person with marshmallow on her head be "h"

consider the motion of the marshmallow after it is dropped from bridge.

Y₀ = initial position of the marshmallow above the ground = 5.71 m

Y = final position of marshmallow on head of person = h

v₀ = initial velocity of the marshmallow = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

t = time of travel for marshmallow = 0.921 sec

Using the kinematics equation

Y = Y₀ + v₀ t + (0.5) a t²

inserting the values

h = 5.71 + 0 (0.921) + (0.5) (-9.8) (0.921)²

h = 5.71 - 4.16

h = 1.55 m

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A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
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v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s

a) The vertical speed when it leaves the ground. is 6.95 m/s

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s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

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If the mass of a material is 50 grams and the volume of the material is 12 cm^3, what would the density of the material be?
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