
A ball thrown horizontally from a point $24m$ above the ground, strikes the ground after traveling horizontally a distance of $18m$ . With what speed was it thrown?
Answer
489.9k+ views
Hint: To solve this type of question we must know concepts of kinematics. Here in this question firstly, we will find the time at which the body travels the distance and then after from the formula of velocity we put the following values to get the required result. And speed is defined as the rate at which distance changes.
Formula used:
$h = ut + \dfrac{1}{2}g{t^2}$
$v = \dfrac{S}{t}$
Where,
$h$ is the height
$u$ is initial velocity,
$S$ is the distance,
$v$ is the velocity.
$g$ is the acceleration due to gravity and
$t$ is the time.
Complete step by step answer:
According to the question we have,
The height from which the ball is thrown $24m$
And the horizontal distance which is travelled by the ball is $18m$
Let us assume that ${u_x}$ be the speed through which the ball is thrown horizontally.
Therefore, the vertical distance travelled in the time $t$ is given by above formula,
i.e.,
$h = ut + \dfrac{1}{2}g{t^2}$
Now substituting all the values in above formula,
$
24 = 0 + \dfrac{1}{2} \times 9.8 \times {t^2} \\
\Rightarrow {t^2} = \dfrac{{2 \times 24}}{{9.8}} \\
\Rightarrow t = \sqrt { = \dfrac{{2 \times 24}}{{9.8}}} \\
\Rightarrow t = 2.21s \\
$
Now we got the time, and now we have to calculate the horizontal component of the velocity which is given by,
$v = \dfrac{S}{t}$
Now substituting all the value in above equation, we get,
$
v = \dfrac{S}{t} \\
\Rightarrow v = \dfrac{{18}}{{2.21}} \\
\Rightarrow v = 8.14m{\text{ }}{s^{ - 1}} \\
$
Hence, the ball is thrown with a speed of $8.14m{\text{ }}{s^{ - 1}}$ .
Note:Note that equations of motion are only valid when there is constant acceleration throughout the motion. Remember to check the unit as all are in standard form. The primary distinction between speed and velocity is that the former has simply magnitude, whilst the latter has both magnitude and direction. It signifies that the velocity is the speed in relation to the direction of travel.
Formula used:
$h = ut + \dfrac{1}{2}g{t^2}$
$v = \dfrac{S}{t}$
Where,
$h$ is the height
$u$ is initial velocity,
$S$ is the distance,
$v$ is the velocity.
$g$ is the acceleration due to gravity and
$t$ is the time.
Complete step by step answer:
According to the question we have,
The height from which the ball is thrown $24m$
And the horizontal distance which is travelled by the ball is $18m$
Let us assume that ${u_x}$ be the speed through which the ball is thrown horizontally.
Therefore, the vertical distance travelled in the time $t$ is given by above formula,
i.e.,
$h = ut + \dfrac{1}{2}g{t^2}$
Now substituting all the values in above formula,
$
24 = 0 + \dfrac{1}{2} \times 9.8 \times {t^2} \\
\Rightarrow {t^2} = \dfrac{{2 \times 24}}{{9.8}} \\
\Rightarrow t = \sqrt { = \dfrac{{2 \times 24}}{{9.8}}} \\
\Rightarrow t = 2.21s \\
$
Now we got the time, and now we have to calculate the horizontal component of the velocity which is given by,
$v = \dfrac{S}{t}$
Now substituting all the value in above equation, we get,
$
v = \dfrac{S}{t} \\
\Rightarrow v = \dfrac{{18}}{{2.21}} \\
\Rightarrow v = 8.14m{\text{ }}{s^{ - 1}} \\
$
Hence, the ball is thrown with a speed of $8.14m{\text{ }}{s^{ - 1}}$ .
Note:Note that equations of motion are only valid when there is constant acceleration throughout the motion. Remember to check the unit as all are in standard form. The primary distinction between speed and velocity is that the former has simply magnitude, whilst the latter has both magnitude and direction. It signifies that the velocity is the speed in relation to the direction of travel.
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