A ball is thrown straight up, from 3 above the ground, with a velocity of 14 /. Gravity
pulls it down. Ignoring air resistance, the distance travelled by the ball with respect to time
can be modelled using the equation ℎ = 3 + 14 − 5
2
, where ℎ is the distance in metres
and is time in seconds.
a) What is the name of the polynomial represented by the above equation?
i) Linear polynomial
ii) Cubic polynomial
iii) Quadratic polynomial
iv) None of the above
b)After how many secondsdoes the ball hit the ground?
i) 0.2
ii) 3
iii) 1
iv) 2
c)Find the maximum height the ball reaches?
i) 10m
ii) 12.8m
iii) 14m
iv) 5m
d) Mention the name of the graph in the given figure
i) Ellipse
ii) Line
iii) Parabola
iv) Circle
e)For an
ℎ degree polynomial, what is the maximum number of zeroes or roots possible?
i) 0
ii) 2n
iii) n
iv) 1
Answers
Step-by-step explanation:
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Falling Object Model
An object is said to be a freely falling body if it is thrown or dropped vertically at an initial velocity and then only gravity affects its motion.
The position of any freely falling body is determined by the initial velocity and the initial height.
If h is the height measured in feet, t is the number of seconds the object has fallen from an initial height h0 with an initial velocity or speed v0 (inft/sec), then the model for height of a falling object is: h(t)=−16t2+v0t+h0
The “ −16t2 ” term comes from the acceleration due to gravity pulling the object towards the ground. The velocity of the object at a particular time t is given by: v(t)=−32t+v0
When an object is thrown upwards from ground with a particular initial velocity, the initial height is zero and when an object is dropped from an initial height the initial velocity is zero. If the value of h is in meters and s in meters/sec, the acceleration due to gravity in meters/sec is 4.9 , the equation becomes: h(t)=−4.9t2+v0t+h0 and then the velocity of the object at a particular time t is given by: v(t)=−9.8t+v0
Example 1:
A ball is thrown vertically upward with an initial speed of 80 ft/sec. How high will the ball be after 3 seconds?
t=3 and v0=80 ft/sec
So:
h=−16(3)2+80(3)
=−144+240
=96 ft
Example 2:
An object is dropped from a height of 120 feet. Assuming that there is no air resistance, how long does it takes to reach the ground?
If h is measured in feet, t is the number of seconds the object had fallen, and h0 is the initial height from which the object was dropped, then the model for the height of falling object is:
h=−16t2+h0
Note that the initial velocity is zero here.
Substitute 0 for h and 120 for h0 in the model. 0=−16t2+120
Solve the equation for t .
0−120=−16t2+120−120
−120=−16t2
−120−16=−16t2−16
7.5=t2
Taking the square root:
t=7.5
≈±2.74
Since t represents time, it cannot be negative.
Therefore, the object will reach the ground in about 2.74 seconds.
These equations are simplified. They ignore air resistance and the gravitational constant is approximate. Also, this model only works for the surface of the Earth (at sea level). The model on other planets will be different because their gravity is different. For example, on the surface of the moon, with h in meters and v0 in m/sec, the falling object model is h=−0.8t2+v0t + h0 .
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