Question

Mustafa pumps a 7.26 kg shot straight up, giving it a constant upward acceleration from rest of 44.0 m/s2 for a height 60.0 cm. He releases it at height 2.30 m above the ground. Ignore air resistance.Calculate the speed of the shot when Mustafa releases it?
How high above the ground does it go?H ow much time does he have to get out of its way before it returns to the height of the top of his head, 1.93 m above the ground?

Answer 1

As Mustafa gives the acceleration of the body within a distance of $S=2.3-60\times 10^{-2}=1.7$ m.

According to the formula $S=\frac{a\times t^2}{2}$ find the travel time: $t=\sqrt{\frac{2\times S} {a}}=\sqrt{\frac{2\times 1.7}{44}} \approx 0.28$ s, hence the velocity $v=a\times t=44\times 0.28=12.32$ m/s is the velocity of the projectile.

According to the law of conservation of energy, the sum of the kinetic and potential energy of the projectile at the time of Mustafa's release is equal to the potential energy at the upper point of the trajectory, that is

$E_C+E_{P1}=E_{P2};$

$\frac{m\times v^2}{2}+m\times h \times g=m\times h_{max}\times g;$

$h_{max}=\frac{v^2+2\times h\times g}{2\times g}=\frac{12.32^2+2\times 2.3\times 10}{2\times 10}\approx 9.89$ m

The fall time to $1.98$ m is $t=\sqrt{\frac{2\times S_1}{g}}=\sqrt{\frac{2\times (9.89-1.93)}{10}}\approx 1.26$ s