Question

A ball thrown by one player reaches the other in 2 sec the maximum height attained by the ball

Answer 1

Answer:

A ball thrown by one player reaches the other in 2 seconds.

so, times of flight , T = 2 sec

Let ball is thrown with speed v which makes an angle \thetaθ with horizontal.

so, horizontal component , v_x=vcos\thetav

x

=vcosθ

and vertical component, v_y=vsin\thetav

y

=vsinθ

total displacement in vertical direction = 0.

so, 0 = v_yT-\frac{1}{2}gT^2v

y

T−

2

1

gT

2

or, v_y(2)=\frac{1}{2}\times10\times(2)^2v

y

(2)=

2

1

×10×(2)

2

or, v_y=10v

y

=10 m/s ......(1)

so, maximum height attained by ball, H = \frac{v^2sin^2\theta}{2g}

2g

v

2

sin

2

θ

= \frac{(vsin\theta)^2}{2g}

2g

(vsinθ)

2

= \frac{v_y^2}{2g}

2g

v

y

2

from equation (1),

= 10²/2(10)

= 5m

hence, maximum height attained by ball is 5m