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Question

When a ball is thrown, its height in feet h after t seconds is given by the equation h=vt-16t^2

where v is the initial upwards velocity in feet per second. If v= 19 feet per second, find all values of t for which h=5 feet. Do not round any intermediate steps. Round your answer to 2 decimal places.

t= ? seconds

I get t=3.8 and -3.8 but I'm not sure that's right.

Answer 1

5=19t-16t^2

16t^2 -19t +5=0

usign quad formula

a=16 b=-19 c 5

19+-√(19^2 -4*16*5)/ 32

19+-√361-320/32

19+-√ 41/32

19+√41/32 or 19-√41/32

0.79 or 0.39

Answer 2

h=vt-16t^2

given, v= 19 ft/sec

h= 5ft  t= ?

5 = 19 t - 16t^2

rearranging the terms

16t^2 -19t +5 = 0

solving this quadratic eqn using _b+-sqrt(b^2 -4ac / 2a

I got t = 0.39sec or t = 0.79 sec

so the answer will be t= 0.39 sec for just going up

and total time will be 0.79 i.e going up and coming down.

and note that time cannot be in negative so answer cant be -3.8

Answer 3

this is the physics dept.

h=vt-16t^2  is  5 =19 * t-16t^2

-16 * t^2 + 19 * t - 5 = 0

values of t are:     0.39365 and   0.79385

Interpretation:

at t= 0.39365 the ball will cross the 5 feet mark and then it will come back at t =  0.79385