Question

1. A ball thrown horizontally from the edge of the top of a building 49 m high strikes the ground 24.5 m from the foot of the building. Find the (a) time it takes the ball to reach the ground, (b) initial velocity of the ball, and (c) velocity just before the ball strikes the ground.

Answer 1

Height of the building = 49 m

Range of the projectile = 24.5 m

A) Time taken by ball :

Time taken by ball to reach the ground is given by;

$\sf:\implies\:t=\sqrt{\dfrac{2H}{g}}$

• H denotes height
• g denotes acceleration due to gravity

$\sf:\implies\:t=\sqrt{\dfrac{2\times 49}{9.8}}$

$\sf:\implies\:t=\sqrt{\dfrac{98}{9.8}}$

$\sf:\implies\:t=\sqrt{10}$

$:\implies\:\underline{\boxed{\bf{\red{t=3.16\:s}}}}$

B) Initial velocity of the ball :

Horizontal distance covered by the ball before hitting the ground is given by;

• Note that here vertical component of the initial velocity will be zero as it is the case of horizontal projectile.
• So the initial will only have horizontal component $\sf{u_x}$

$\sf:\implies\:R=u_x\times t$

$\sf:\implies\:24.5=u_x\times 3.16$

$\sf:\implies\:u_x=\dfrac{24.5}{3.16}$

$:\implies\:\underline{\boxed{\bf{\blue{u_x=7.75\:ms^{-1}}}}}$

C) Final velocity of the ball :

Velocity with which ball strikes the ground will have two components. i.e, Horizontal and vertical (downward).

• Horizontal component $\sf{v_x}$
• Vertical component $\sf{v_y}$

$\sf:\implies\:v=\sqrt{{v_x}^2+{v_y}^2}$

• Horizontal component of the velocity will remain same.

$\sf:\implies\:v=\sqrt{{u_x}^2+{gt}^2}$

$\sf:\implies\:v=\sqrt{{7.75}^2+{(10\times 3.16)}^2}$

$\sf:\implies\:v=\sqrt{60+998.56}$

$\sf:\implies\:v=\sqrt{1058.56}$

$:\implies\:\underline{\boxed{\bf{\gray{v=32.53\:ms^{-1}}}}}$

Answer 2

Solution :

• Height of the building = 49 m
• Range = 24.5 m
• We are asked to find the time it takes for thr ball to reach the ground & the velocity of the ball just before it strikes the ground

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According to the Question :

• The time taken is given by t = √(2H/g)

⇒ t = √(2H/g)

• H = 49 m
• g = 9.8 m/s²

⇒ t = √[(2×49)/9.8]

⇒ t = √(98/9.8)

⇒ t = √10 seonds

• The initial velocity can be found with the help of the relation R = ut

⇒ R = ut

• R = 24.5 m
• t = √10 sec

⇒ 24.5 = u × √10

⇒ (24.5)/(√10) = u

⇒ u = 7.75 m/s

• The velocity of the ball just before it strikes the groumd is given by the magnitude of its components

⇒ v = √vx +vy

• vx = 10 m/s
• vy = √2gh = √2×10×49 = √980 m/s

⇒ v = √(100 + 980)

⇒ v = √(1080)

⇒ v = 32.8 m/s

Therefore the time taken by the ball is √10 sec & the initial velocity of the ball is 7.75 m/s & the velocity of the ball just before it strikes the ground is 32.8 m/s