A ball is thrown vertically upward from at a height of 5m above the ground with

Question

A ball is thrown vertically upward from at a height of 5m above the ground with a velocity of 25 m/s. a) How long will it take to rise to its highest point? b) How high does the ball rise from the ground? c) How long after projection will the ball have a velocity of 10m/s upward? d) How long after projection will the ball have a velocity of 10m/s downward? e) How long will it take to hit the ground after it was thrown? 2. Given vectors A=3i + 2j-5k, B = -i + 3j-2k & C=2i-4j + 6k Find: a)A-2B+C b)A middot (B times C) C)AX(B times C) d) The angle between vectors A and B, A B =| A | | B | cos theta 3. A subway train starts from rest at a station and accelerates at a rate of 2m/s^2 for 10 seconds. It then runs at the constant speed for 30s, and decelerates at 4m/s until it stops at the next station. Find the total distance covered.

Explanation / Answer

1) u = 25 m/s, h = 5m

a) t = u/g =25/9.8 = 2.55 s

b) H = h + u^2/2g = (25*25)/(2*9.8) =36.89 m

c) v = u -gt

10-25 =-9.8*t

t = 1.53 s

d) -10-25 = -9.8*t

t =3.57 s

e) time of ascent t1 = 2.55s

final veocity v = [2gH]^1/2 = [2*9.8*36.89]^1/2 = 26.89 m/s

time of decent t2 = v/g = 2.744s

Total time = t1 +t2 = 2.55 + 2.744 =5.294 s

2) a) A-2B+C = (3i +2j -5k)-2(-i+3j-2k)+(2i -4j+6k) = 7i-8j+5k

b) A.(BxC)

A.(BxC) = (3i +2j -5k).[(-i+3j-2k)x (2i -4j+6k)= 44

c) cos(theta) = A.B/A*B = (-3+6+10)/(9+4+25)^1/2(1+9+4)^1/2

Angle between A and B is = 55.7 degrees

3) a =2 m/s^s , t =10s

s1 = ut+(1/2)at^2 = 0+(0.5*2*10*10) = 100 m

v =u+at = 0+2*10 = 20 m/s

s2 = v*t = 20*30 = 600 m

v^2 -u^2 = 2as

s3 = (20*20)/(2*4) = 50 m

Total distance S = s1+s2+s3 = 100+600+ 50

s = 750 m

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