The acceleration due to Earth’s gravity is 32 ft/sec2. From ground level, a projectile is fired straight upward with velocity 90 feet per second. When does (in secs) it hit the ground?
A. 6.25
B. 5.625
C. 11.25
D. 12.5
Answer
Verified
479.1k+ views
Hint: When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.
Complete step by step solution:
A projectile is fired straight upward with the velocity 90 feet per second at the acceleration due to Earth’s gravity of 32 ft/sec2.
$v(t)=v(0)+(-32t)$
To find when it hits the ground, find the time when it was at peak and double it.
Its peaks at
$\begin{align}
& v(t)=0 \\
& v(0)=90\text{ (given)} \\
\end{align}$
Place the known values –
$\begin{align}
& v(t)=v(0)+(-32t) \\
& 0=90-32t \\
\end{align}$
Take negative terms on the left side of the equation.
$32t=90$
[When the multiplicative term changes its side, it goes in the denominator]
$\begin{align}
& \therefore t=\dfrac{90}{32} \\
& \therefore t=2.8125 \\
\end{align}$
Object hits the ground at time,
$\begin{align}
& t=2.8125\times 2 \\
& t=5.625\text{ Sec} \\
\end{align}$
Hence, from the given multiple options, option B is the correct answer.
Additional information: As an object falls, its speed increases because it’s being pulled on by gravity. The acceleration of gravity near the earth is $g=-9.81{m}/{{{s}^{2}}}\;$ to find out the speed or the velocity after a certain amount of time, you just multiply the acceleration of gravity by the amount of time since it was let go of. So, you get: velocity $v=-9.81{m}/{{{s}^{2}}\times time}\;$ or V = gt. The negative sign just means that the object is moving downwards. If it were positive, then it would be moving up. For speed rather than velocity, you just drop the negative sign.
If you have an initial velocity (if you threw the ball up or down instead of just letting go of it), then you have to include this in the equation, too, giving you: V = Vo + gt, where Vo is the initial velocity of the object. This equation will still work if you threw the ball to the side, instead of straight up or down, except that it will only give you the up-down velocity, not the total velocity. (And the number you should use for Vo is still just the up-down velocity that the object starts with.)
Note: The acceleration of a body in free fall under the influence of earth's gravity expressed as the rate of increase of the velocity per unit of time and assigned the standard value of 980.665 centimeters per second per second are also called g.
Complete step by step solution:
A projectile is fired straight upward with the velocity 90 feet per second at the acceleration due to Earth’s gravity of 32 ft/sec2.
$v(t)=v(0)+(-32t)$
To find when it hits the ground, find the time when it was at peak and double it.
Its peaks at
$\begin{align}
& v(t)=0 \\
& v(0)=90\text{ (given)} \\
\end{align}$
Place the known values –
$\begin{align}
& v(t)=v(0)+(-32t) \\
& 0=90-32t \\
\end{align}$
Take negative terms on the left side of the equation.
$32t=90$
[When the multiplicative term changes its side, it goes in the denominator]
$\begin{align}
& \therefore t=\dfrac{90}{32} \\
& \therefore t=2.8125 \\
\end{align}$
Object hits the ground at time,
$\begin{align}
& t=2.8125\times 2 \\
& t=5.625\text{ Sec} \\
\end{align}$
Hence, from the given multiple options, option B is the correct answer.
Additional information: As an object falls, its speed increases because it’s being pulled on by gravity. The acceleration of gravity near the earth is $g=-9.81{m}/{{{s}^{2}}}\;$ to find out the speed or the velocity after a certain amount of time, you just multiply the acceleration of gravity by the amount of time since it was let go of. So, you get: velocity $v=-9.81{m}/{{{s}^{2}}\times time}\;$ or V = gt. The negative sign just means that the object is moving downwards. If it were positive, then it would be moving up. For speed rather than velocity, you just drop the negative sign.
If you have an initial velocity (if you threw the ball up or down instead of just letting go of it), then you have to include this in the equation, too, giving you: V = Vo + gt, where Vo is the initial velocity of the object. This equation will still work if you threw the ball to the side, instead of straight up or down, except that it will only give you the up-down velocity, not the total velocity. (And the number you should use for Vo is still just the up-down velocity that the object starts with.)
Note: The acceleration of a body in free fall under the influence of earth's gravity expressed as the rate of increase of the velocity per unit of time and assigned the standard value of 980.665 centimeters per second per second are also called g.
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