A toy rocket (not internally powered) is launched straight up from the top of a

Wierzycaz

Wierzycaz

Answered question

2021-09-24

A toy rocket (not internally powered) is launched straight up from the top of a building 50 ft tall at an initial velocity of 200 ft per sec.
a) Give the function that describes the height of the rocket in terms of time t.
b) Determine the time at which the rocket reaches its maximum height and the maximum height in feet.
c) For what time interval will be rocket be more than 300 ft above ground level?
d) After how many seconds will it hit the ground?

Answer & Explanation

Arham Warner

Arham Warner

Skilled2021-09-25Added 102 answers

Step 1
The first three questions receive standard responses. to investigate how the thing moves under the current circumstances
Step 2
Let H(t) represent the height of the object above the ground. H(t) is calculated using the common equations of motion (under gravity), as indicated. Recall that at t=0, the height is 50 (as given)
a) H(t)=ut+0.5at2+50
=200t16t2+50
Step 3b) filling in the square to determine the highest point (the same outcome may be attained by settingtting H(t)=20032t=0)
b) Write
H(t)=200t16t2+50
=5016(t6.25)2+16×6.252
=67516(t6.25)2
So, H(t)=maximum at t=6.25sec 
Step 4
c) a height above the earth of at least 300 feet throughout the period between 1.41 and 11.1 seconds
c) Necessary
H(t)=200t16t2+50300
solve 200t16t2+50=300
16t2200t+250=0
t1.41,11.1 So,
height H(t)300 between 1.41 and 11.1 sec

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