Question

The ball leaves the bat at the angle of 25 degrees with a Velocity of 30 m/s.

Questions:
a. How long will it takes for the ball to hit the ground?
b. How high will it go?
c. At what distance does the ball covers?

Solution:
x = Vox t
y = Viy t + ½gt 2
t = Viy/(g)

Answer 1

Answer:

The time taken by the ball to reach the highest point is half the total time of flight. As the time of ascending and descending is same for a projectile without air resistance. the time to reach the highest point T

t

H

=

2

T

=

g

usinθ

=

10

30

×sin30

o

=1.5s

b. The maximum height reached is

2g

u

2

sin

2

θ

=

2g

(30

2

)×(sin30

o

)

2

=

2×10×4

900

=11.25m

c. Horizontal range =

g

u

2

sin2θ

=

10

(30)

2

sin2(30

o

)

=

20

900

3

=45

3

m

d. The time for which thc ball is in air is same as its time of eight, i.e.,

g

2usinθ

=

10

2×30×sin30

o

=3s