A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground. (c) the velocity (direction and magnitude) with which it strikes the ground.Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics
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This case is a projectile fired horizontally from a height of h
g=9.8m/s2
a)Time it takes to reach the ground,
height=100m
t=√2h/g
t=√2x100/9.8=√200/9.8=4.51 sec
b)The horizontal distance it travels before reaching the ground:
speed=u=20m/s
x=ut=20x 4.5 =90m
c)Horizontal velocity remains constant through out the journey.
At "A", vx=20m/s
vy=u+at
=0+9.8 x 4.5=44.1m/s
Final resultant velocity=vr=√(44.1)²+20²=48.42 m/s
tanβ=vy/vx=44.1/20=2.205
β=tan⁻¹(2.205)=66⁰
∴The ball strikes the ground with a velocity of 48.42m/s at an angle of 66° with horizontal.
This case is a projectile fired horizontally from a height of h
g=9.8m/s2
a)Time it takes to reach the ground,
height=100m
t=√2h/g
t=√2x100/9.8=√200/9.8=4.51 sec
b)The horizontal distance it travels before reaching the ground:
speed=u=20m/s
x=ut=20x 4.5 =90m
c)Horizontal velocity remains constant through out the journey.
At "A", vx=20m/s
vy=u+at
=0+9.8 x 4.5=44.1m/s
Final resultant velocity=vr=√(44.1)²+20²=48.42 m/s
tanβ=vy/vx=44.1/20=2.205
β=tan⁻¹(2.205)=66⁰
∴The ball strikes the ground with a velocity of 48.42m/s at an angle of 66° with horizontal.
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