1.

A ball is thrown vertically upwards from the top of a tower with an initial velocity of 39.2 m/s. Theball reaches the ground after 5 s. Calculate (i) the height of the tower (ii) the velocity of ball on reaching the ground. (take g=9.8 m/s2)​

Answer»

S=ut+
2
1

at
2

=19.6×5−
2
1

×9.8×5
2

=−24.5 m

Hence, HEIGHT of TOWER =24.5m
v=u+at
=19.6−9.8×5=−29.4 m/s



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