Question

(a) A projectile is fired with a velocity v0 making an angle θ0 with the horizontal. Derive

expressions for (i) time of flight (ii) maximum height and (iii) horizontal range of the

projectile.

(b)A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much

high above the ground can the cricketer throw the same ball?

OR

(a)Derive an expression for the magnitude of the acceleration of a particle moving with

uniform speed v along a circular arc of radius r. Show that it is directed along the radius

towards the center of the circular arc.

(b)A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in the

x-y plane with a constant acceleration of a = 8.0 i + 2.0 j ms-2

. At what time is the x-

coordinate of the particle 16 m? What is the speed of the particle at that time?​

Answer 1

a) 2) maximum height.

b)

Medium

Updated on : 2022-09-05

Solution

verified

Verified by Toppr

Maximum horizontal distance, R=100m

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45

∘

, i.e., =45

∘

.

The max horizontal range for a projection velocity v is given by the relation:

R

max

=

g

u

2

100=

g

u

2

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.

Acceleration, a=g

Using the third equation of motion:

v

2

−u

2

=−2gH

H=u

2

/2g=100/2=50m