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Standard XII

Physics

Acceleration

A man throws ...

Question

A man throws a ball at height of 10 m at an angle of $35^{o}$ from horizontal. If mass of ball is 0.5 kg and its initial speed is 30 m/s.
$g = - 9.8 \frac{m}{s^{2}}$
How long is the ball in the air?

2.26s

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4.02s

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1.76s

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5.05s

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1.12s

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Solution

The correct option is A 4.02s
The speed of ball in vertically upwards direction= $30 \times s i n 35^{\circ} m / s = 17.21 m / s$
The net displacement is $= - 10 m$
Using the equation of motion $s = u t + \frac{1}{2} a t^{2}$
$⟹ - 10 = (30 s i n 35^{\circ}) t - \frac{1}{2} g t^{2}$
$⟹ t = 4.02 s$

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A man throws a ball at height of 10 m at an angle of 35o from horizontal. If mass of ball is 0.5 kg and its initial speed is 30 m/s.g=−9.8ms2How long is the ball in the air?

The correct option is A 4.02sThe speed of ball in vertically upwards direction=30×sin35∘m/s=17.21m/sThe net displacement is =−10mUsing the equation of motion s=ut+12at2⟹−10=(30sin35∘)t−12gt2⟹t=4.02s

A man throws a ball at height of 10 m at an angle of $35^{o}$ from horizontal. If mass of ball is 0.5 kg and its initial speed is 30 m/s.
$g = - 9.8 \frac{m}{s^{2}}$
How long is the ball in the air?