A ball is dropped from a platform 19.6 m high. Its position function is -

Question

A ball is dropped from a platform 19.6 m high. Its position function is - (A) x = - 4.9t2 + 19.6 (0 le t le 1) (B) x = - 4.9t2 + 19.6 (0 le t le 2) (C) x = - 9.8t2 + 19.6 (0 le t le 2) (D) x = - 4.9t2 - 19.6 (0 le t le 2)

Tardigrade Admin · Accepted Answer

We have, a=(d2 x/d t2)=-9.8 The initial conditions are x(0)=19.6 and v(0)=0 So, v=(d x/d t)=-9.8 t+v(0)=-9.8 t ∴ x=-4.9 t2+x(0)=-4.9 t2+19.6 Now, the domain of the function is restricted since the ball hits the ground after a certain time. To find this time we set x=0 and solve for t. 0=4.9 t2+19.6 ⇒ t=2

Q. A ball is dropped from a platform $19.6 m$ high. Its position function is -

$x = - 4.9 t^{2} + 19.6 (0 \leq t \leq 1)$

$x = - 4.9 t^{2} + 19.6 (0 \leq t \leq 2)$

$x = - 9.8 t^{2} + 19.6 (0 \leq t \leq 2)$

$x = - 4.9 t^{2} - 19.6 (0 \leq t \leq 2)$

Q. A ball is dropped from a platform 19.6m high. Its position function is -

x=−4.9t2+19.6(0≤t≤1)

x=−4.9t2+19.6(0≤t≤2)

x=−9.8t2+19.6(0≤t≤2)

x=−4.9t2−19.6(0≤t≤2)

Q. A ball is dropped from a platform $19.6 m$ high. Its position function is -

$x = - 4.9 t^{2} + 19.6 (0 \leq t \leq 1)$

$x = - 4.9 t^{2} + 19.6 (0 \leq t \leq 2)$

$x = - 9.8 t^{2} + 19.6 (0 \leq t \leq 2)$

$x = - 4.9 t^{2} - 19.6 (0 \leq t \leq 2)$