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Standard XII

Physics

Graph of Variation of G

If an object ...

Question

If an object is thrown straight up into the air with an initial velocity of $112$ feet per second, its height after t second is given by the formula : $h = 112 t - 16 t^{2}$ , where h represents the height of the object in feet. After the object has been thrown, in how many seconds will it hit the ground?

7

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Solution

The correct option is B $7$
$h = 112 t - {16 t}^{2}$
At ground $h = 0$
$∴$ $112 t - {16 t}^{2} = 0 \Rightarrow 16 t = 112$
$t = 7 s$
Therefore after $7$ seconds object hits the ground.

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If an object is thrown straight up into the air with an initial velocity of 112 feet per second, its height after t second is given by the formula : h=112t−16t2, where h represents the height of the object in feet. After the object has been thrown, in how many seconds will it hit the ground?

The correct option is B 7h=112t−16t2At ground h=0∴ 112t−16t2=0⇒16t=112t=7sTherefore after 7 seconds object hits the ground.

The correct option is B $7$
$h = 112 t - {16 t}^{2}$
At ground $h = 0$
$∴$ $112 t - {16 t}^{2} = 0 \Rightarrow 16 t = 112$
$t = 7 s$
Therefore after $7$ seconds object hits the ground.