Topic :- Types of projectile inotion
19. 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 (di-
rection and magnitude) with which it strikes the ground.
Ans. (a) 4.5$ (b) 90 m
(c) 49 m/s. 066° with horizontal
Answers
Answer:
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★Given:-
- A ball is thrown horizontally from a point above the ground.
- Height - 100m
- Speed = 20m/s.
★To 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.
★Solution:-
Given that A ball is thrown horizontally from a point above the ground.Hence this is a case of projectile fired horizontally from a height.
a)Time it takes to reach the ground:-
Using the formula,
✦t=√2h/g
Where we have,
- t = time
- h = height = 100m
- g = 9.8
Putting values,
➜t=√2x100/9.8
➜t = √200/9.8
➜t = 4.51 s
Therefore,the time taken to reach the ground is 4.5s.
b)The horizontal distance it travels before reaching the ground:-
Using the formula,
✦Distance = speed×time
Putting values,
➜Distance = ut
➜20x 4.5
➜90m
Therefore,the horizontal distance it travels before reaching the ground is 90m.
c)Velocity with which it strikes the ground:-
Through out the journey,the horizontal velocity remains constant as it is a projectile.Therefore,
➜ Vₓ = 20m/s
➜Vᵧ = u+at
➜0+9.8 x 4.5
➜44.1m/s
Final resultant velocity:-
➜Vᵣ = √(44.1)²+20²
=48.42 m/s
➜tanβ = Vᵧ/Vₓ
=44.1/20
=2.205
➜β=tan⁻¹(2.205)=66⁰.
Therefore,the ball strikes the ground with a velocity 48.42 m/s at an angle 66° with horizontal.
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