Question

A catapult hurls a stone of mass 32.0 g with velocity of 50.0 m/s at a 30.0 degree angle of elevation.

a) What is the maximum height reached by the stone?
b) What is its range ?
c) How long has the stone been in the air when it returns to its original height? ​

Answer 1

ANSWER

S

y

=ut−

2

1

gt

2

h=S

y

=20sin30

o

×5−

2

1

×10×(5)

2

h=50−125=−75m (minus sign just indicates that the displacement is in downward direction)

h

max

=

2g

u

2

sin

2

θ

=

2×10

(20)

2

sin

2

30

0

=5m

Hence, maximum height attained by projectile =h+h

max

=75+5=80 m

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Answer 2

Hello friend!

Your answer is in attachment