f(t)=−4.9t2+50t+5 
f(0)=5f(4)=−4.9∗42+50∗4+5=126.6 
A)average velocity = 4−0f(4)−f(0) =121.6/4=30.4 m/s 
B) instantaneous velocity = f′(4) =−2∗4.9∗t+50=−2∗4.9∗4+50=10.8m/s 
C) Acceleration =f′′(t)=d(f′(t))/dt=−2∗4.9 
f′′(4)=−2∗4.9=−9.8m/s2  
D) Maximum height achieved by the tomato = maximum value of f(t) 
 f′(t)=0−2∗4.9∗t+50=0t=25/4.9 sec.t=5.1 secf(5.1)=−4.9∗(5.1)2+50∗5.1+5=132.5m 
E)When an object is thrown vertically upwards then time taken by the object to achieve the maximum height is equal to time taken by the object to return to its initial position from the maximum height   
Time the tomato is in air is twice the time taken by the tomato to achieve the maximum height 
Time = 2*5.1 =10.2 sec.
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