45 degrees below the ground II

A projectile is thrown horizontally from a 20 m 20\text{ m} building. If, at the landing point, the velocity vector of the projectile makes an angle of 4 5 45^\circ below the ground, calculate the projectile's initial speed in m/s \text{m/s} .

Take g g (the accelaration due to gravity) as 10 m/s 2 10 \text{ m/s}^2 .


The answer is 20.

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3 solutions

Rishabh Tiwari
Jun 14, 2016

Since the projectile is thrown horizontally, therefore the initial y c o m p o n e n t \color{#20A900}{y-component} of its velocity is zero ;

Therefore at any height ' h \color{#3D99F6}{h} ' ,

V y V_y = = 2 g h \sqrt {2gh} ...[from 3 r d 3^{rd} e q . n eq.^{n} of motion]

N o w \color{#3D99F6}{Now} ; At the instant of landing,

V y V x \dfrac {V_y}{V_x} = = tan 4 5 \tan45^{\circ}

\Rightarrow V y V_y = = V x V_x

\Rightarrow V x V_x = = 2 g h \sqrt {2gh}

Plugging in g g = = 10 10 \text {&} h h = = 20 20 ;

we get \rightarrow

V x = 20 \color{#69047E}{\boxed {V_x = 20}}

As the problem says that ball is thrown horizontally from a height of 20m .If we consider 20 m as height of projectile and use H= u^2 sin^2(theta)/2(g) and plug in H=20m and theta =45 degrees and g=10m/s^2, we get u= 20(1.414)= 28.2 m/s . Which is not an answer.

CAN SOMEBODY PLEASE POINT OUT THE BUG IN MY SOLUTION ?

Saraswati Sharma - 4 years, 8 months ago

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The answer you ended up with is the final speed, but the initial speed is only equal to the horizontal component of the velocity, because it has not undergone acceleration in the y direction until it has started to fall.

Tristan Goodman - 2 years, 2 months ago
Erasmo Hinojosa
Jun 4, 2016

At the landing point, the magnitud of the vertical and horizontal components of the velocity should be the same since the velocity vector makes and angle of 45 degrees with respect to the ground. The initial velocity was purely horizontal and since that component of the velocity remained unchanged through all the flight, the initial speed equals the magnitud of the vertical component of the projectile's velocity at the landing point.

The vertical position of the projectile, measured from the ground, is given by: y = h 1 2 g t 2 y=h- \frac{1}{2} g t^2 Where "h" is the buildings height.

And its vertical component of the velocity is given by: v y = g t v_y=-gt

Using the first equation for solving for the time of the projectile´s flight (y=0m at landing) and using the secong one for solving for the vertical component of the velocity and having in mind that: v 0 = v y \|v_0\|=\|v_y\| where v 0 v_0 is the projectile's initial velocity.

Finally, this yields: v 0 = 2 h g \|v_0\|=\sqrt{2hg} and by replacing values for "h" and "g" we obtain that v 0 = 20 m s \|v_0\|=20 \frac{m}{s} So, the answer is "20".

Nice explanation to a nice question, + 1 ! \color{#D61F06}{+1!}

Rishabh Tiwari - 5 years ago
N. Aadhaar Murty
Oct 4, 2020

The projectile takes 2 H g = 2 s \frac {2H}{g} = 2 s to reach the ground. Since the final velocity vector forms an angle of 4 5 45^{\circ} with the ground, V x = V y V_x = V_y . V x = 2 g = 20 \therefore V_x = 2g = 20

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